МенюИнтеграли, съдържащи $\sqrt{x^2 - a^2}$
$\displaystyle \int \frac{dx}{\sqrt{x^2 - a^2}} = \ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int \frac{x\,dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2}$
$\displaystyle \int \frac{x^2\,dx}{\sqrt{x^2 - a^2}} = \frac{x\sqrt{x^2 - a^2}}{2} + \frac{a^2}{2}\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int \frac{x^3\,dx}{\sqrt{x^2 - a^2}} = \frac{(x^2 - a^2)^{3/2}}{3} + a^2\sqrt{x^2 - a^2}$
$\displaystyle \int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a}\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \frac{dx}{x^2\sqrt{x^2 - a^2}} = \frac{\sqrt{x^2 - a^2}}{a^2x}$
$\displaystyle \int \frac{dx}{x^3\sqrt{x^2 - a^2}} = \frac{\sqrt{x^2 - a^2}}{2a^2x^2} + \frac{1}{2a^3}\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \sqrt{x^2 - a^2}\,dx = \frac{x\sqrt{x^2 - a^2}}{2} - \frac{a^2}{2}\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int x\sqrt{x^2 - a^2}\,dx = \frac{(x^2 - a^2)^{3/2}}{3}$
$\displaystyle \int x^2\sqrt{x^2 - a^2}\,dx = \frac{x(x^2 - a^2)^{3/2}}{4} + \frac{a^2x\sqrt{x^2 - a^2}}{8} - \frac{a^4}{8}\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int x^3\sqrt{x^2 - a^2}\,dx = \frac{(x^2 - a^2)^{5/2}}{5} + \frac{a^2(x^2 - a^2)^{3/2}}{3}$
$\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x}\,dx = \sqrt{x^2 - a^2} - a\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x^2}\,dx = -\frac{\sqrt{x^2 - a^2}}{x} + \ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x^3}\,dx = -\frac{\sqrt{x^2 - a^2}}{2x^2} + \frac{1}{2a}\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \frac{dx}{(x^2 - a^2)^{3/2}} = -\frac{x}{a^2\sqrt{x^2 - a^2}}$
$\displaystyle \int \frac{x\,dx}{(x^2 - a^2)^{3/2}} = \frac{-1}{\sqrt{x^2 - a^2}}$
$\displaystyle \int \frac{x^2\,dx}{(x^2 - a^2)^{3/2}} = -\frac{x}{\sqrt{x^2 - a^2}} + \ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int \frac{x^3\,dx}{(x^2 - a^2)^{3/2}} = \sqrt{x^2 - a^2} - \frac{a^2}{\sqrt{x^2 - a^2}}$
$\displaystyle \int \frac{dx}{x(x^2 - a^2)^{3/2}} = \frac{-1}{a^2\sqrt{x^2 - a^2}} - \frac{1}{a^3}\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \frac{dx}{x^2(x^2 - a^2)^{3/2}} = -\frac{\sqrt{x^2 - a^2}}{a^4x} - \frac{x}{a^4\sqrt{x^2 - a^2}}$
$\displaystyle \int \frac{dx}{x^3(x^2 - a^2)^{3/2}} = \frac{1}{2a^2x^2\sqrt{x^2 - a^2}} - \frac{3}{2a^4\sqrt{x^2 - a^2}} - \frac{3}{2a^5}\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int (x^2 -\nobreak a^2)^{3/2}\,dx = \frac{x(x^2 - a^2)^{3/2}}{4} - \frac{3a^2x\sqrt{x^2 - a^2}}{8} + \frac{3}{8}a^4\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int x(x^2 -\nobreak a^2)^{3/2}\,dx = \frac{(x^2 - a^2)^{5/2}}{5}$
$\displaystyle \int x^2(x^2 -\nobreak a^2)^{3/2}\,dx = \frac{x(x^2 - a^2)^{5/2}}{6} + \frac{a^2x(x^2 - a^2)^{3/2}}{24} - \frac{a^4x\sqrt{x^2 - a^2}}{16} + \frac{a^6}{16}\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int x^3(x^2 -\nobreak a^2)^{3/2}\,dx = \frac{(x^2 - a^2)^{7/2}}{7} + \frac{a^2(x^2 - a^2)^{5/2}}{5}$
$\displaystyle \int \frac{(x^2 - a^2)^{3/2}}{x}\,dx = \frac{(x^2 - a^2)^{3/2}}{3} - a^2\sqrt{x^2 - a^2} + a^3\sec^{-1}\left|\frac{x}{a}\right|$
$\displaystyle \int \frac{(x^2 - a^2)^{3/2}}{x^2}\,dx = -\frac{(x^2 - a^2)^{3/2}}{x} + \frac{3x\sqrt{x^2 - a^2}}{2} - \frac{3}{2}a^2\ln\left(x + \sqrt{x^2 - a^2}\right)$
$\displaystyle \int \frac{(x^2 - a^2)^{3/2}}{x^3}\,dx = -\frac{(x^2 - a^2)^{3/2}}{2x^2} + \frac{3\sqrt{x^2 - a^2}}{2} - \frac{3}{2}a\sec^{-1}\left|\frac{x}{a}\right|$
Интеграли, съдържащи $\sqrt{a^2 - x^2}$
$\displaystyle \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{\sqrt{a^2 - x^2}} = -\sqrt{a^2 - x^2}$
$\displaystyle \int \frac{x^2\,dx}{\sqrt{a^2 - x^2}} = -\frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x^3\,dx}{\sqrt{a^2 - x^2}} = \frac{(a^2 - x^2)^{3/2}}{3} - a^2\sqrt{a^2 - x^2}$
$\displaystyle \int \frac{dx}{x\sqrt{a^2 - x^2}} = -\frac{1}{a}\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \frac{dx}{x^2\sqrt{a^2 - x^2}} = -\frac{\sqrt{a^2 - x^2}}{a^2x}$
$\displaystyle \int \frac{dx}{x^3\sqrt{a^2 - x^2}} = -\frac{\sqrt{a^2 - x^2}}{2a^2x^2} - \frac{1}{2a^3}\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \sqrt{a^2 - x^2}\,dx = \frac{x\sqrt{a^2 - x^2}}{2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}$
$\displaystyle \int x\sqrt{a^2 - x^2}\,dx = -\frac{(a^2 - x^2)^{3/2}}{3}$
$\displaystyle \int x^2\sqrt{a^2 - x^2}\,dx = -\frac{x(a^2 - x^2)^{3/2}}{4} + \frac{a^2x\sqrt{a^2 - x^2}}{8} + \frac{a^4}{8}\sin^{-1}\frac{x}{a}$
$\displaystyle \int x^3\sqrt{a^2 - x^2}\,dx = \frac{(a^2 - x^2)^{5/2}}{5} - \frac{a^2(a^2 - x^2)^{3/2}}{3}$
$\displaystyle \int \frac{\sqrt{a^2 - x^2}}{x}\,dx = \sqrt{a^2 - x^2} - a\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \frac{\sqrt{a^2 - x^2}}{x^2}\,dx = -\frac{\sqrt{a^2 - x^2}}{x} - \sin^{-1}\frac{x}{a}$
$\displaystyle \int \frac{\sqrt{a^2 - x^2}}{x^3}\,dx = -\frac{\sqrt{a^2 - x^2}}{2x^2} + \frac{1}{2a}\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \frac{dx}{(a^2 - x^2)^{3/2}} = \frac{x}{a^2\sqrt{a^2 - x^2}}$
$\displaystyle \int \frac{x\,dx}{(a^2 - x^2)^{3/2}} = \frac{1}{\sqrt{a^2 - x^2}}$
$\displaystyle \int \frac{x^2\,dx}{(a^2 - x^2)^{3/2}} = \frac{x}{\sqrt{a^2 - x^2}} - \sin^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x^3\,dx}{(a^2 - x^2)^{3/2}} = \sqrt{a^2 - x^2} + \frac{a^2}{\sqrt{a^2 - x^2}}$
$\displaystyle \int \frac{dx}{x(a^2 - x^2)^{3/2}} = \frac{1}{a^2\sqrt{a^2 - x^2}} - \frac{1}{a^3}\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \frac{dx}{x^2(a^2 - x^2)^{3/2}} = -\frac{\sqrt{a^2 - x^2}}{a^4x} + \frac{x}{a^4\sqrt{a^2 - x^2}}$
$\displaystyle \int \frac{dx}{x^3(a^2 - x^2)^{3/2}} = \frac{-1}{2a^2x^2\sqrt{a^2 - x^2}} + \frac{3}{2a^4\sqrt{a^2 - x^2}} - \frac{3}{2a^5}\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int (a^2 -\nobreak x^2)^{3/2}\,dx = \frac{x(a^2 - x^2)^{3/2}}{4} + \frac{3a^2x\sqrt{a^2 - x^2}}{8} + \frac{3}{8}a^4\sin^{-1}\frac{x}{a}$
$\displaystyle \int x(a^2 -\nobreak x^2)^{3/2}\,dx = -\frac{(a^2 - x^2)^{5/2}}{5}$
$\displaystyle \int x^2(a^2 -\nobreak x^2)^{3/2}\,dx = -\frac{x(a^2 - x^2)^{5/2}}{6} + \frac{a^2x(a^2 - x^2)^{3/2}}{24} + \frac{a^4x\sqrt{a^2 - x^2}}{16} + \frac{a^6}{16}\sin^{-1}\frac{x}{a}$
$\displaystyle \int x^3(a^2 -\nobreak x^2)^{3/2}\,dx = \frac{(a^2 - x^2)^{7/2}}{7} - \frac{a^2(a^2 - x^2)^{5/2}}{5}$
$\displaystyle \int \frac{(a^2 - x^2)^{3/2}}{x}\,dx = \frac{(a^2 - x^2)^{3/2}}{3} + a^2\sqrt{a^2 - x^2} - a^3\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
$\displaystyle \int \frac{(a^2 - x^2)^{3/2}}{x^2}\,dx = -\frac{(a^2 - x^2)^{3/2}}{x} - \frac{3x\sqrt{a^2 - x^2}}{2} - \frac{3}{2}a^2\sin^{-1}\frac{x}{a}$
$\displaystyle \int \frac{(a^2 - x^2)^{3/2}}{x^3}\,dx = -\frac{(a^2 - x^2)^{3/2}}{2x^2} - \frac{3\sqrt{a^2 - x^2}}{2} + \frac{3}{2}a\ln\left(\frac{a + \sqrt{a^2 - x^2}}{x}\right)$
Интеграли, съдържащи $ax^2 + bx + c$
$\displaystyle \int \frac{dx}{ax^2 + bx + c} = \frac{2}{\sqrt{4ac - b^2}}\tan^{-1}\frac{2ax + b}{\sqrt{4ac - b^2}}$ $[b^2 \lt 4ac]$
$\displaystyle \int \frac{dx}{ax^2 + bx + c} = \frac{1}{\sqrt{b^2 - 4ac}}\,\allowbreak\ln\left(\frac{2ax + b - \sqrt{b^2 - 4ac}}{2ax + b + \sqrt{b^2 - 4ac}}\right)$ $[b^2 > 4ac]$
Ако $b^2 = 4ac$, то $ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2$.
$\displaystyle \int \frac{x\,dx}{ax^2 + bx + c} = \frac{1}{2a}\ln(ax^2 +\nobreak bx +\nobreak c) - \frac{b}{2a}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{x^2\,dx}{ax^2 + bx + c} = \frac{x}{a} - \frac{b}{2a^2}\ln(ax^2 +\nobreak bx +\nobreak c) + \frac{b^2 - 2ac}{2a^2}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{x^m\,dx}{ax^2 + bx + c} = \frac{x^{m-1}}{(m - 1)a} - \frac{c}{a}\int \frac{x^{m-2}\,dx}{ax^2 + bx + c} - \frac{b}{a}\int \frac{x^{m-1}\,dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{dx}{x(ax^2 + bx + c)} = \frac{1}{2c}\ln\left(\frac{x^2}{ax^2 + bx + c}\right) - \frac{b}{2c}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{dx}{x^2(ax^2 + bx + c)} = \frac{b}{2c^2}\ln\left(\frac{ax^2 + bx + c}{x^2}\right) - \frac{1}{cx} + \frac{b^2 - 2ac}{2c^2}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{dx}{x^n(ax^2 + bx + c)} = -\frac{1}{(n - 1)cx^{n-1}} - \frac{b}{c}\int \frac{dx}{x^{n-1}(ax^2 + bx + c)} - \frac{a}{c}\int \frac{dx}{x^{n-2}(ax^2 + bx + c)}$
$\displaystyle \int \frac{dx}{(ax^2 + bx + c)^2} = \frac{2ax + b}{(4ac - b^2)(ax^2 + bx + c)} + \frac{2a}{4ac - b^2}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{x\,dx}{(ax^2 + bx + c)^2} = -\frac{bx + 2c}{(4ac - b^2)(ax^2 + bx + c)} - \frac{b}{4ac - b^2}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{x^2\,dx}{(ax^2 + bx + c)^2} = \frac{(b^2 - 2ac)x + bc}{a(4ac - b^2)(ax^2 + bx + c)} + \frac{2c}{4ac - b^2}\int \frac{dx}{ax^2 + bx + c}$
$\displaystyle \int \frac{x^m\,dx}{(ax^2 + bx + c)^n} = -\frac{x^{m-1}}{(2n - m - 1)a(ax^2 + bx + c)^{n-1}} + \frac{(m - 1)c}{(2n - m - 1)a}\,\allowbreak\int \frac{x^{m-2}\,dx}{(ax^2 + bx + c)^n} - \frac{(n - m)b}{(2n - m - 1)a}\,\allowbreak\int \frac{x^{m-1}\,dx}{(ax^2 + bx + c)^n}$
$\displaystyle \int \frac{x^{2n-1}\,dx}{(ax^2 + bx + c)^n} = \frac{1}{a}\int \frac{x^{2n-3}\,dx}{(ax^2 + bx + c)^{n-1}} - \frac{c}{a}\int \frac{x^{2n-3}\,dx}{(ax^2 + bx + c)^n} - \frac{b}{a}\int \frac{x^{2n-2}\,dx}{(ax^2 + bx + c)^n}$
$\displaystyle \int \frac{dx}{x(ax^2 + bx + c)^2} = \frac{1}{2c(ax^2 + bx + c)} - \frac{b}{2c}\int \frac{dx}{(ax^2 + bx + c)^2} + \frac{1}{c}\int \frac{dx}{x(ax^2 + bx + c)}$
$\displaystyle \int \frac{dx}{x^2(ax^2 + bx + c)^2} = -\frac{1}{cx(ax^2 + bx + c)} - \frac{3a}{c}\int \frac{dx}{(ax^2 + bx + c)^2} - \frac{2b}{c}\int \frac{dx}{x(ax^2 + bx + c)^2}$
$\displaystyle \int \frac{dx}{x^m(ax^2 + bx + c)^n} = -\frac{1}{(m - 1)cx^{m-1}(ax^2 + bx + c)^{n-1}} - \frac{(m + 2n - 3)a}{(m - 1)c}\,\allowbreak\int \frac{dx}{x^{m-2}(ax^2 + bx + c)^n} - \frac{(m + n - 2)b}{(m - 1)c}\,\allowbreak\int \frac{dx}{x^{m-1}(ax^2 + bx + c)^n}$
Интеграли, съдържащи $\sqrt{ax^2 + bx + c}$
В долните формули, ако $b^2 = 4ac$, то $\sqrt{ax^2 + bx + c} = \sqrt{a}\left(x + \frac{b}{2a}\right)$.
$\displaystyle \int \frac{dx}{\sqrt{ax^2 + bx + c}} = \frac{1}{\sqrt{a}}\,\allowbreak\ln\left(2\sqrt{a}\sqrt{ax^2 + bx + c} + 2ax + b\right)$ $[a > 0]$
$\displaystyle \int \frac{dx}{\sqrt{ax^2 + bx + c}} = -\frac{1}{\sqrt{-a}}\sin^{-1}\left(\frac{2ax + b}{\sqrt{b^2 - 4ac}}\right)$ $[a \lt 0]$
$\displaystyle \int \frac{dx}{\sqrt{ax^2 + bx + c}} = \frac{1}{\sqrt{a}}\sinh^{-1}\left(\frac{2ax + b}{\sqrt{4ac - b^2}}\right)$ $[a > 0,\ b^2 \lt 4ac]$
$\displaystyle \int \frac{x\,dx}{\sqrt{ax^2 + bx + c}} = \frac{\sqrt{ax^2 + bx + c}}{a} - \frac{b}{2a}\int \frac{dx}{\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{x^2\,dx}{\sqrt{ax^2 + bx + c}} = \frac{2ax - 3b}{4a^2}\sqrt{ax^2 + bx + c} + \frac{3b^2 - 4ac}{8a^2}\int \frac{dx}{\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{dx}{x\sqrt{ax^2 + bx + c}} = -\frac{1}{\sqrt{c}}\,\allowbreak\ln\left(\frac{2\sqrt{c}\sqrt{ax^2 + bx + c} + bx + 2c}{x}\right)$ $[c > 0]$
$\displaystyle \int \frac{dx}{x\sqrt{ax^2 + bx + c}} = \frac{1}{\sqrt{-c}}\sin^{-1}\left(\frac{bx + 2c}{|x|\sqrt{b^2 - 4ac}}\right)$ $[c \lt 0]$
$\displaystyle \int \frac{dx}{x\sqrt{ax^2 + bx + c}} = -\frac{1}{\sqrt{c}}\sinh^{-1}\left(\frac{bx + 2c}{|x|\sqrt{4ac - b^2}}\right)$ $[c > 0,\ b^2 \lt 4ac]$
$\displaystyle \int \frac{dx}{x^2\sqrt{ax^2 + bx + c}} = -\frac{\sqrt{ax^2 + bx + c}}{cx} - \frac{b}{2c}\int \frac{dx}{x\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \sqrt{ax^2 + bx + c}\,dx = \frac{(2ax + b)\sqrt{ax^2 + bx + c}}{4a} + \frac{4ac - b^2}{8a}\int \frac{dx}{\sqrt{ax^2 + bx + c}}$
$\displaystyle \int x\sqrt{ax^2 + bx + c}\,dx = \frac{(ax^2 + bx + c)^{3/2}}{3a} - \frac{b(2ax + b)}{8a^2}\sqrt{ax^2 + bx + c} - \frac{b(4ac - b^2)}{16a^2}\int \frac{dx}{\sqrt{ax^2 + bx + c}}$
$\displaystyle \int x^2\sqrt{ax^2 + bx + c}\,dx = \frac{6ax - 5b}{24a^2}(ax^2 +\nobreak bx +\nobreak c)^{3/2} + \frac{5b^2 - 4ac}{16a^2}\int \sqrt{ax^2 + bx + c}\,dx$
$\displaystyle \int \frac{\sqrt{ax^2 + bx + c}}{x}\,dx = \sqrt{ax^2 + bx + c} + \frac{b}{2}\int \frac{dx}{\sqrt{ax^2 + bx + c}} + c\int \frac{dx}{x\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{\sqrt{ax^2 + bx + c}}{x^2}\,dx = -\frac{\sqrt{ax^2 + bx + c}}{x} + a\int \frac{dx}{\sqrt{ax^2 + bx + c}} + \frac{b}{2}\int \frac{dx}{x\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{dx}{(ax^2 + bx + c)^{3/2}} = \frac{2(2ax + b)}{(4ac - b^2)\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{x\,dx}{(ax^2 + bx + c)^{3/2}} = \frac{2(bx + 2c)}{(b^2 - 4ac)\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{x^2\,dx}{(ax^2 + bx + c)^{3/2}} = \frac{(2b^2 - 4ac)x + 2bc}{a(4ac - b^2)\sqrt{ax^2 + bx + c}} + \frac{1}{a}\int \frac{dx}{\sqrt{ax^2 + bx + c}}$
$\displaystyle \int \frac{dx}{x(ax^2 + bx + c)^{3/2}} = \frac{1}{c\sqrt{ax^2 + bx + c}} + \frac{1}{c}\int \frac{dx}{x\sqrt{ax^2 + bx + c}} - \frac{b}{2c}\int \frac{dx}{(ax^2 + bx + c)^{3/2}}$
$\displaystyle \int \frac{dx}{x^2(ax^2 + bx + c)^{3/2}} = -\frac{ax^2 + 2bx + c}{c^2x\sqrt{ax^2 + bx + c}} + \frac{b^2 - 2ac}{2c^2}\int \frac{dx}{(ax^2 + bx + c)^{3/2}} - \frac{3b}{2c^2}\int \frac{dx}{x\sqrt{ax^2 + bx + c}}$
$\displaystyle \int (ax^2 +\nobreak bx +\nobreak c)^{n + 1/2}\,dx = \frac{(2ax + b)(ax^2 + bx + c)^{n + 1/2}}{4a(n + 1)} + \frac{(2n + 1)(4ac - b^2)}{8a(n + 1)}\,\allowbreak\int (ax^2 +\nobreak bx +\nobreak c)^{n - 1/2}\,dx$
$\displaystyle \int x(ax^2 +\nobreak bx +\nobreak c)^{n + 1/2}\,dx = \frac{(ax^2 + bx + c)^{n + 3/2}}{a(2n + 3)} - \frac{b}{2a}\int (ax^2 +\nobreak bx +\nobreak c)^{n + 1/2}\,dx$
$\displaystyle \int \frac{dx}{(ax^2 + bx + c)^{n + 1/2}} = \frac{2(2ax + b)}{(2n - 1)(4ac - b^2)(ax^2 + bx + c)^{n - 1/2}} + \frac{8a(n - 1)}{(2n - 1)(4ac - b^2)}\,\allowbreak\int \frac{dx}{(ax^2 + bx + c)^{n - 1/2}}$
$\displaystyle \int \frac{dx}{x(ax^2 + bx + c)^{n + 1/2}} = \frac{1}{(2n - 1)c(ax^2 + bx + c)^{n - 1/2}} + \frac{1}{c}\int \frac{dx}{x(ax^2 + bx + c)^{n - 1/2}} - \frac{b}{2c}\int \frac{dx}{(ax^2 + bx + c)^{n + 1/2}}$
Интеграли, съдържащи $x^3 + a^3$
За формулите, съдържащи $x^3 - a^3$, заменете $a$ с $-a$.
$\displaystyle \int \frac{dx}{x^3 + a^3} = \frac{1}{6a^2}\ln\frac{(x + a)^2}{x^2 - ax + a^2} + \frac{1}{a^2\sqrt{3}}\tan^{-1}\frac{2x - a}{a\sqrt{3}}$
$\displaystyle \int \frac{x\,dx}{x^3 + a^3} = \frac{1}{6a}\ln\frac{x^2 - ax + a^2}{(x + a)^2} + \frac{1}{a\sqrt{3}}\tan^{-1}\frac{2x - a}{a\sqrt{3}}$
$\displaystyle \int \frac{x^2\,dx}{x^3 + a^3} = \frac{1}{3}\ln(x^3 +\nobreak a^3)$
$\displaystyle \int \frac{dx}{x(x^3 + a^3)} = \frac{1}{3a^3}\ln\left(\frac{x^3}{x^3 + a^3}\right)$
$\displaystyle \int \frac{dx}{x^2(x^3 + a^3)} = -\frac{1}{a^3x} - \frac{1}{6a^4}\ln\frac{x^2 - ax + a^2}{(x + a)^2} - \frac{1}{a^4\sqrt{3}}\tan^{-1}\frac{2x - a}{a\sqrt{3}}$
$\displaystyle \int \frac{dx}{(x^3 + a^3)^2} = \frac{x}{3a^3(x^3 + a^3)} + \frac{1}{9a^5}\ln\frac{(x + a)^2}{x^2 - ax + a^2} + \frac{2}{3a^5\sqrt{3}}\tan^{-1}\frac{2x - a}{a\sqrt{3}}$
$\displaystyle \int \frac{x\,dx}{(x^3 + a^3)^2} = \frac{x^2}{3a^3(x^3 + a^3)} + \frac{1}{18a^4}\ln\frac{x^2 - ax + a^2}{(x + a)^2} + \frac{1}{3a^4\sqrt{3}}\tan^{-1}\frac{2x - a}{a\sqrt{3}}$
$\displaystyle \int \frac{x^2\,dx}{(x^3 + a^3)^2} = -\frac{1}{3(x^3 + a^3)}$
$\displaystyle \int \frac{dx}{x(x^3 + a^3)^2} = \frac{1}{3a^3(x^3 + a^3)} + \frac{1}{3a^6}\ln\left(\frac{x^3}{x^3 + a^3}\right)$
$\displaystyle \int \frac{dx}{x^2(x^3 + a^3)^2} = -\frac{1}{a^6x} - \frac{x^2}{3a^6(x^3 + a^3)} - \frac{4}{3a^6}\int \frac{x\,dx}{x^3 + a^3}$
$\displaystyle \int \frac{x^m\,dx}{x^3 + a^3} = \frac{x^{m-2}}{m - 2} - a^3\int \frac{x^{m-3}\,dx}{x^3 + a^3}$
$\displaystyle \int \frac{dx}{x^n(x^3 + a^3)} = \frac{-1}{a^3(n - 1)x^{n-1}} - \frac{1}{a^3}\int \frac{dx}{x^{n-3}(x^3 + a^3)}$
Интеграли, съдържащи $x^4 \pm a^4$
$\displaystyle \int \frac{dx}{x^4 + a^4} = \frac{1}{4a^3\sqrt{2}}\ln\left(\frac{x^2 + ax\sqrt{2} + a^2}{x^2 - ax\sqrt{2} + a^2}\right) - \frac{1}{2a^3\sqrt{2}}\tan^{-1}\frac{ax\sqrt{2}}{x^2 - a^2}$
$\displaystyle \int \frac{x\,dx}{x^4 + a^4} = \frac{1}{2a^2}\tan^{-1}\frac{x^2}{a^2}$
$\displaystyle \int \frac{x^2\,dx}{x^4 + a^4} = \frac{1}{4a\sqrt{2}}\ln\left(\frac{x^2 - ax\sqrt{2} + a^2}{x^2 + ax\sqrt{2} + a^2}\right) - \frac{1}{2a\sqrt{2}}\tan^{-1}\frac{ax\sqrt{2}}{x^2 - a^2}$
$\displaystyle \int \frac{x^3\,dx}{x^4 + a^4} = \frac{1}{4}\ln(x^4 +\nobreak a^4)$
$\displaystyle \int \frac{dx}{x(x^4 + a^4)} = \frac{1}{4a^4}\ln\left(\frac{x^4}{x^4 + a^4}\right)$
$\displaystyle \int \frac{dx}{x^2(x^4 + a^4)} = -\frac{1}{a^4x} - \frac{1}{4a^5\sqrt{2}}\ln\left(\frac{x^2 - ax\sqrt{2} + a^2}{x^2 + ax\sqrt{2} + a^2}\right) + \frac{1}{2a^5\sqrt{2}}\tan^{-1}\frac{ax\sqrt{2}}{x^2 - a^2}$
$\displaystyle \int \frac{dx}{x^3(x^4 + a^4)} = -\frac{1}{2a^4x^2} - \frac{1}{2a^6}\tan^{-1}\frac{x^2}{a^2}$
$\displaystyle \int \frac{dx}{x^4 - a^4} = \frac{1}{4a^3}\ln\left(\frac{x - a}{x + a}\right) - \frac{1}{2a^3}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{x^4 - a^4} = \frac{1}{4a^2}\ln\left(\frac{x^2 - a^2}{x^2 + a^2}\right)$
$\displaystyle \int \frac{x^2\,dx}{x^4 - a^4} = \frac{1}{4a}\ln\left(\frac{x - a}{x + a}\right) + \frac{1}{2a}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x^3\,dx}{x^4 - a^4} = \frac{1}{4}\ln(x^4 -\nobreak a^4)$
$\displaystyle \int \frac{dx}{x(x^4 - a^4)} = \frac{1}{4a^4}\ln\left(\frac{x^4 - a^4}{x^4}\right)$
$\displaystyle \int \frac{dx}{x^2(x^4 - a^4)} = \frac{1}{a^4x} + \frac{1}{4a^5}\ln\left(\frac{x - a}{x + a}\right) + \frac{1}{2a^5}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{dx}{x^3(x^4 - a^4)} = \frac{1}{2a^4x^2} + \frac{1}{4a^6}\ln\left(\frac{x^2 - a^2}{x^2 + a^2}\right)$
Интеграли, съдържащи $x^n \pm a^n$
$\displaystyle \int \frac{dx}{x(x^n + a^n)} = \frac{1}{na^n}\ln\frac{x^n}{x^n + a^n}$
$\displaystyle \int \frac{x^{n-1}\,dx}{x^n + a^n} = \frac{1}{n}\ln(x^n +\nobreak a^n)$
$\displaystyle \int \frac{x^m\,dx}{(x^n + a^n)^r} = \int \frac{x^{m-n}\,dx}{(x^n + a^n)^{r-1}} - a^n\int \frac{x^{m-n}\,dx}{(x^n + a^n)^r}$
$\displaystyle \int \frac{dx}{x^m(x^n + a^n)^r} = \frac{1}{a^n}\int \frac{dx}{x^m(x^n + a^n)^{r-1}} - \frac{1}{a^n}\int \frac{dx}{x^{m-n}(x^n + a^n)^r}$
$\displaystyle \int \frac{dx}{x\sqrt{x^n + a^n}} = \frac{1}{n\sqrt{a^n}}\ln\left(\frac{\sqrt{x^n + a^n} - \sqrt{a^n}}{\sqrt{x^n + a^n} + \sqrt{a^n}}\right)$
$\displaystyle \int \frac{dx}{x(x^n - a^n)} = \frac{1}{na^n}\ln\left(\frac{x^n - a^n}{x^n}\right)$
$\displaystyle \int \frac{x^{n-1}\,dx}{x^n - a^n} = \frac{1}{n}\ln(x^n -\nobreak a^n)$
$\displaystyle \int \frac{x^m\,dx}{(x^n - a^n)^r} = a^n\int \frac{x^{m-n}\,dx}{(x^n - a^n)^r} + \int \frac{x^{m-n}\,dx}{(x^n - a^n)^{r-1}}$
$\displaystyle \int \frac{dx}{x^m(x^n - a^n)^r} = \frac{1}{a^n}\int \frac{dx}{x^{m-n}(x^n - a^n)^r} - \frac{1}{a^n}\int \frac{dx}{x^m(x^n - a^n)^{r-1}}$
$\displaystyle \int \frac{dx}{x\sqrt{x^n - a^n}} = \frac{2}{n\sqrt{a^n}}\cos^{-1}\sqrt{\frac{a^n}{x^n}}$
$\displaystyle \int \frac{x^{p-1}\,dx}{x^{2m} + a^{2m}} = \frac{1}{ma^{2m-p}}\sum_{k=1}^{m}\sin\frac{(2k - 1)p\pi}{2m}\,\allowbreak\tan^{-1}\left(\frac{x - a\cos[(2k - 1)\pi/2m]}{a\sin[(2k - 1)\pi/2m]}\right) - \frac{1}{2ma^{2m-p}}\sum_{k=1}^{m}\cos\frac{(2k - 1)p\pi}{2m}\,\allowbreak\ln\left(x^2 - 2ax\cos\frac{(2k - 1)\pi}{2m} + a^2\right)$ [където $0 \lt p \le 2m$]
$\displaystyle \int \frac{x^{p-1}\,dx}{x^{2m} - a^{2m}} = \frac{1}{2ma^{2m-p}}\sum_{k=1}^{m-1}\cos\frac{kp\pi}{m}\,\allowbreak\ln\left(x^2 - 2ax\cos\frac{k\pi}{m} + a^2\right) - \frac{1}{ma^{2m-p}}\sum_{k=1}^{m-1}\sin\frac{kp\pi}{m}\,\allowbreak\tan^{-1}\left(\frac{x - a\cos(k\pi/m)}{a\sin(k\pi/m)}\right) + \frac{1}{2ma^{2m-p}}\,\allowbreak\left\{\ln(x -\nobreak a) + (-1)^p\ln(x +\nobreak a)\right\}$ [където $0 \lt p \le 2m$]
$\displaystyle \int \frac{x^{p-1}\,dx}{x^{2m+1} + a^{2m+1}} = \frac{2(-1)^{p-1}}{(2m + 1)a^{2m-p+1}}\sum_{k=1}^{m}\sin\frac{2kp\pi}{2m + 1}\,\allowbreak\tan^{-1}\left(\frac{x + a\cos[2k\pi/(2m + 1)]}{a\sin[2k\pi/(2m + 1)]}\right) + \frac{(-1)^{p-1}}{(2m + 1)a^{2m-p+1}}\sum_{k=1}^{m}\cos\frac{2kp\pi}{2m + 1}\,\allowbreak\ln\left(x^2 + 2ax\cos\frac{2k\pi}{2m + 1} + a^2\right) + \frac{(-1)^{p-1}\ln(x + a)}{(2m + 1)a^{2m-p+1}}$ [където $0 \lt p \le 2m + 1$]
$\displaystyle \int \frac{x^{p-1}\,dx}{x^{2m+1} - a^{2m+1}} = \frac{-2}{(2m + 1)a^{2m-p+1}}\sum_{k=1}^{m}\sin\frac{2kp\pi}{2m + 1}\,\allowbreak\tan^{-1}\left(\frac{x - a\cos[2k\pi/(2m + 1)]}{a\sin[2k\pi/(2m + 1)]}\right) + \frac{1}{(2m + 1)a^{2m-p+1}}\sum_{k=1}^{m}\cos\frac{2kp\pi}{2m + 1}\,\allowbreak\ln\left(x^2 - 2ax\cos\frac{2k\pi}{2m + 1} + a^2\right) + \frac{\ln(x - a)}{(2m + 1)a^{2m-p+1}}$ [където $0 \lt p \le 2m + 1$]
Интеграли, съдържащи $\sin ax$
$\displaystyle \int \sin ax\,dx = -\frac{\cos ax}{a}$
$\displaystyle \int x\sin ax\,dx = \frac{\sin ax}{a^2} - \frac{x\cos ax}{a}$
$\displaystyle \int x^2\sin ax\,dx = \frac{2x\sin ax}{a^2} + \left(\frac{2}{a^3} - \frac{x^2}{a}\right)\cos ax$
$\displaystyle \int x^3\sin ax\,dx = \left(\frac{3x^2}{a^2} - \frac{6}{a^4}\right)\sin ax + \left(\frac{6x}{a^3} - \frac{x^3}{a}\right)\cos ax$
$\displaystyle \int \frac{\sin ax}{x}\,dx = ax - \frac{(ax)^3}{3\cdot 3!} + \frac{(ax)^5}{5\cdot 5!} - \cdots$
$\displaystyle \int \frac{\sin ax}{x^2}\,dx = -\frac{\sin ax}{x} + a\int \frac{\cos ax}{x}\,dx$
$\displaystyle \int \frac{dx}{\sin ax} = \frac{1}{a}\ln(\csc ax -\nobreak \cot ax) = \frac{1}{a}\ln\tan\frac{ax}{2}$
$\displaystyle \int \frac{x\,dx}{\sin ax} = \frac{1}{a^2}\bigg\{ax + \frac{(ax)^3}{18} + \frac{7(ax)^5}{1800} + \cdots + \frac{2(2^{2n-1} - 1)B_n(ax)^{2n+1}}{(2n + 1)!} + \cdots\bigg\}$
Тук $B_n$ са числата на Бернули: $B_1 = \frac{1}{6}$, $B_2 = \frac{1}{30}$, $B_3 = \frac{1}{42}$, $B_4 = \frac{1}{30}$, $B_5 = \frac{5}{66}, \ldots$
$\displaystyle \int \sin^2 ax\,dx = \frac{x}{2} - \frac{\sin 2ax}{4a}$
$\displaystyle \int x\sin^2 ax\,dx = \frac{x^2}{4} - \frac{x\sin 2ax}{4a} - \frac{\cos 2ax}{8a^2}$
$\displaystyle \int \sin^3 ax\,dx = -\frac{\cos ax}{a} + \frac{\cos^3 ax}{3a}$
$\displaystyle \int \sin^4 ax\,dx = \frac{3x}{8} - \frac{\sin 2ax}{4a} + \frac{\sin 4ax}{32a}$
$\displaystyle \int \frac{dx}{\sin^2 ax} = -\frac{\cot ax}{a}$
$\displaystyle \int \frac{dx}{\sin^3 ax} = -\frac{\cos ax}{2a\sin^2 ax} + \frac{1}{2a}\ln\tan\frac{ax}{2}$
$\displaystyle \int \sin px\sin qx\,dx = \frac{\sin(p - q)x}{2(p - q)} - \frac{\sin(p + q)x}{2(p + q)}$ [при $p = \pm q$, вж. $\int \sin^2 ax\,dx$]
$\displaystyle \int \frac{dx}{1 - \sin ax} = \frac{1}{a}\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right)$
$\displaystyle \int \frac{x\,dx}{1 - \sin ax} = \frac{x}{a}\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right) + \frac{2}{a^2}\ln\sin\left(\frac{\pi}{4} - \frac{ax}{2}\right)$
$\displaystyle \int \frac{dx}{1 + \sin ax} = -\frac{1}{a}\tan\left(\frac{\pi}{4} - \frac{ax}{2}\right)$
$\displaystyle \int \frac{x\,dx}{1 + \sin ax} = -\frac{x}{a}\tan\left(\frac{\pi}{4} - \frac{ax}{2}\right) + \frac{2}{a^2}\ln\sin\left(\frac{\pi}{4} + \frac{ax}{2}\right)$
$\displaystyle \int \frac{dx}{(1 - \sin ax)^2} = \frac{1}{2a}\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right) + \frac{1}{6a}\tan^3\left(\frac{\pi}{4} + \frac{ax}{2}\right)$
$\displaystyle \int \frac{dx}{(1 + \sin ax)^2} = -\frac{1}{2a}\tan\left(\frac{\pi}{4} - \frac{ax}{2}\right) - \frac{1}{6a}\tan^3\left(\frac{\pi}{4} - \frac{ax}{2}\right)$
$\displaystyle \int \frac{dx}{p + q\sin ax} = \frac{2}{a\sqrt{p^2 - q^2}}\tan^{-1}\frac{p\tan\frac{1}{2}ax + q}{\sqrt{p^2 - q^2}}$ $[p^2 > q^2]$
$\displaystyle \int \frac{dx}{p + q\sin ax} = \frac{1}{a\sqrt{q^2 - p^2}}\,\allowbreak\ln\left(\frac{p\tan\frac{1}{2}ax + q - \sqrt{q^2 - p^2}}{p\tan\frac{1}{2}ax + q + \sqrt{q^2 - p^2}}\right)$ $[p^2 \lt q^2]$
При $p = \pm q$ вж. $\int \frac{dx}{1 \pm \sin ax}$.
$\displaystyle \int \frac{dx}{(p + q\sin ax)^2} = \frac{q\cos ax}{a(p^2 - q^2)(p + q\sin ax)} + \frac{p}{p^2 - q^2}\int \frac{dx}{p + q\sin ax}$
При $p = \pm q$ вж. $\int \frac{dx}{(1 \pm \sin ax)^2}$.
$\displaystyle \int \frac{dx}{p^2 + q^2\sin^2 ax} = \frac{1}{ap\sqrt{p^2 + q^2}}\tan^{-1}\frac{\sqrt{p^2 + q^2}\tan ax}{p}$
$\displaystyle \int \frac{dx}{p^2 - q^2\sin^2 ax} = \frac{1}{ap\sqrt{p^2 - q^2}}\tan^{-1}\frac{\sqrt{p^2 - q^2}\tan ax}{p}$ $[p^2 > q^2]$
$\displaystyle \int \frac{dx}{p^2 - q^2\sin^2 ax} = \frac{1}{2ap\sqrt{q^2 - p^2}}\,\allowbreak\ln\left(\frac{\sqrt{q^2 - p^2}\tan ax + p}{\sqrt{q^2 - p^2}\tan ax - p}\right)$ $[p^2 \lt q^2]$
$\displaystyle \int x^m\sin ax\,dx = -\frac{x^m\cos ax}{a} + \frac{mx^{m-1}\sin ax}{a^2} - \frac{m(m - 1)}{a^2}\int x^{m-2}\sin ax\,dx$
$\displaystyle \int \frac{\sin ax}{x^n}\,dx = -\frac{\sin ax}{(n - 1)x^{n-1}} + \frac{a}{n - 1}\int \frac{\cos ax}{x^{n-1}}\,dx$ [вж. $\int \frac{\cos ax}{x^n}\,dx$]
$\displaystyle \int \sin^n ax\,dx = -\frac{\sin^{n-1} ax\cos ax}{an} + \frac{n - 1}{n}\int \sin^{n-2} ax\,dx$
$\displaystyle \int \frac{dx}{\sin^n ax} = \frac{-\cos ax}{a(n - 1)\sin^{n-1} ax} + \frac{n - 2}{n - 1}\int \frac{dx}{\sin^{n-2} ax}$
$\displaystyle \int \frac{x\,dx}{\sin^n ax} = \frac{-x\cos ax}{a(n - 1)\sin^{n-1} ax} - \frac{1}{a^2(n - 1)(n - 2)\sin^{n-2} ax} + \frac{n - 2}{n - 1}\int \frac{x\,dx}{\sin^{n-2} ax}$
Интеграли, съдържащи $\cos ax$
$\displaystyle \int \cos ax\,dx = \frac{\sin ax}{a}$
$\displaystyle \int x\cos ax\,dx = \frac{\cos ax}{a^2} + \frac{x\sin ax}{a}$
$\displaystyle \int x^2\cos ax\,dx = \frac{2x\cos ax}{a^2} + \left(\frac{x^2}{a} - \frac{2}{a^3}\right)\sin ax$
$\displaystyle \int x^3\cos ax\,dx = \left(\frac{3x^2}{a^2} - \frac{6}{a^4}\right)\cos ax + \left(\frac{x^3}{a} - \frac{6x}{a^3}\right)\sin ax$
$\displaystyle \int \frac{\cos ax}{x}\,dx = \ln x - \frac{(ax)^2}{2\cdot 2!} + \frac{(ax)^4}{4\cdot 4!} - \cdots$
$\displaystyle \int \frac{\cos ax}{x^2}\,dx = -\frac{\cos ax}{x} - a\int \frac{\sin ax}{x}\,dx$
$\displaystyle \int \frac{dx}{\cos ax} = \frac{1}{a}\ln(\sec ax +\nobreak \tan ax) = \frac{1}{a}\ln\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right)$
$\displaystyle \int \frac{x\,dx}{\cos ax} = \frac{1}{a^2}\bigg\{\frac{(ax)^2}{2} + \frac{(ax)^4}{8} + \frac{5(ax)^6}{144} + \cdots + \frac{E_n(ax)^{2n+2}}{(2n + 2)(2n)!} + \cdots\bigg\}$
Тук $E_n$ са числата на Ойлер: $E_0 = 1$, $E_1 = 1$, $E_2 = 5$, $E_3 = 61$, $E_4 = 1385, \ldots$
$\displaystyle \int \cos^2 ax\,dx = \frac{x}{2} + \frac{\sin 2ax}{4a}$
$\displaystyle \int x\cos^2 ax\,dx = \frac{x^2}{4} + \frac{x\sin 2ax}{4a} + \frac{\cos 2ax}{8a^2}$
$\displaystyle \int \cos^3 ax\,dx = \frac{\sin ax}{a} - \frac{\sin^3 ax}{3a}$
$\displaystyle \int \cos^4 ax\,dx = \frac{3x}{8} + \frac{\sin 2ax}{4a} + \frac{\sin 4ax}{32a}$
$\displaystyle \int \frac{dx}{\cos^2 ax} = \frac{\tan ax}{a}$
$\displaystyle \int \frac{dx}{\cos^3 ax} = \frac{\sin ax}{2a\cos^2 ax} + \frac{1}{2a}\ln\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right)$
$\displaystyle \int \cos ax\cos px\,dx = \frac{\sin(a - p)x}{2(a - p)} + \frac{\sin(a + p)x}{2(a + p)}$ [при $a = \pm p$, вж. $\int \cos^2 ax\,dx$]
$\displaystyle \int \frac{dx}{1 - \cos ax} = -\frac{1}{a}\cot\frac{ax}{2}$
$\displaystyle \int \frac{x\,dx}{1 - \cos ax} = -\frac{x}{a}\cot\frac{ax}{2} + \frac{2}{a^2}\ln\sin\frac{ax}{2}$
$\displaystyle \int \frac{dx}{1 + \cos ax} = \frac{1}{a}\tan\frac{ax}{2}$
$\displaystyle \int \frac{x\,dx}{1 + \cos ax} = \frac{x}{a}\tan\frac{ax}{2} + \frac{2}{a^2}\ln\cos\frac{ax}{2}$
$\displaystyle \int \frac{dx}{(1 - \cos ax)^2} = -\frac{1}{2a}\cot\frac{ax}{2} - \frac{1}{6a}\cot^3\frac{ax}{2}$
$\displaystyle \int \frac{dx}{(1 + \cos ax)^2} = \frac{1}{2a}\tan\frac{ax}{2} + \frac{1}{6a}\tan^3\frac{ax}{2}$
$\displaystyle \int \frac{dx}{p + q\cos ax} = \frac{2}{a\sqrt{p^2 - q^2}}\,\allowbreak\tan^{-1}\left(\sqrt{(p - q)/(p + q)}\,\tan\tfrac{1}{2}ax\right)$ $[p^2 > q^2]$
$\displaystyle \int \frac{dx}{p + q\cos ax} = \frac{1}{a\sqrt{q^2 - p^2}}\,\allowbreak\ln\left(\frac{\tan\frac{1}{2}ax + \sqrt{(q + p)/(q - p)}}{\tan\frac{1}{2}ax - \sqrt{(q + p)/(q - p)}}\right)$ $[p^2 \lt q^2]$
При $p = \pm q$ вж. $\int \frac{dx}{1 \pm \cos ax}$.
$\displaystyle \int \frac{dx}{(p + q\cos ax)^2} = \frac{q\sin ax}{a(q^2 - p^2)(p + q\cos ax)} - \frac{p}{q^2 - p^2}\int \frac{dx}{p + q\cos ax}$
При $p = \pm q$ вж. $\int \frac{dx}{(1 \pm \cos ax)^2}$.
$\displaystyle \int \frac{dx}{p^2 + q^2\cos^2 ax} = \frac{1}{ap\sqrt{p^2 + q^2}}\tan^{-1}\frac{p\tan ax}{\sqrt{p^2 + q^2}}$
$\displaystyle \int \frac{dx}{p^2 - q^2\cos^2 ax} = \frac{1}{ap\sqrt{p^2 - q^2}}\tan^{-1}\frac{p\tan ax}{\sqrt{p^2 - q^2}}$ $[p^2 > q^2]$
$\displaystyle \int \frac{dx}{p^2 - q^2\cos^2 ax} = \frac{1}{2ap\sqrt{q^2 - p^2}}\,\allowbreak\ln\left(\frac{p\tan ax - \sqrt{q^2 - p^2}}{p\tan ax + \sqrt{q^2 - p^2}}\right)$ $[p^2 \lt q^2]$
$\displaystyle \int x^m\cos ax\,dx = \frac{x^m\sin ax}{a} + \frac{mx^{m-1}}{a^2}\cos ax - \frac{m(m - 1)}{a^2}\int x^{m-2}\cos ax\,dx$
$\displaystyle \int \frac{\cos ax}{x^n}\,dx = -\frac{\cos ax}{(n - 1)x^{n-1}} - \frac{a}{n - 1}\int \frac{\sin ax}{x^{n-1}}\,dx$ [вж. $\int \frac{\sin ax}{x^n}\,dx$]
$\displaystyle \int \cos^n ax\,dx = \frac{\sin ax\cos^{n-1} ax}{an} + \frac{n - 1}{n}\int \cos^{n-2} ax\,dx$
$\displaystyle \int \frac{dx}{\cos^n ax} = \frac{\sin ax}{a(n - 1)\cos^{n-1} ax} + \frac{n - 2}{n - 1}\int \frac{dx}{\cos^{n-2} ax}$
$\displaystyle \int \frac{x\,dx}{\cos^n ax} = \frac{x\sin ax}{a(n - 1)\cos^{n-1} ax} - \frac{1}{a^2(n - 1)(n - 2)\cos^{n-2} ax} + \frac{n - 2}{n - 1}\int \frac{x\,dx}{\cos^{n-2} ax}$
Интеграли, съдържащи $\sin ax$ и $\cos ax$
$\displaystyle \int \sin ax\cos ax\,dx = \frac{\sin^2 ax}{2a}$
$\displaystyle \int \sin px\cos qx\,dx = -\frac{\cos(p - q)x}{2(p - q)} - \frac{\cos(p + q)x}{2(p + q)}$
$\displaystyle \int \sin^n ax\cos ax\,dx = \frac{\sin^{n+1} ax}{(n + 1)a}$ [при $n = -1$: $\frac{1}{a}\ln\sin ax$]
$\displaystyle \int \cos^n ax\sin ax\,dx = -\frac{\cos^{n+1} ax}{(n + 1)a}$ [при $n = -1$: $-\frac{1}{a}\ln\cos ax$]
$\displaystyle \int \sin^2 ax\cos^2 ax\,dx = \frac{x}{8} - \frac{\sin 4ax}{32a}$
$\displaystyle \int \frac{dx}{\sin ax\cos ax} = \frac{1}{a}\ln\tan ax$
$\displaystyle \int \frac{dx}{\sin^2 ax\cos ax} = \frac{1}{a}\ln\tan\left(\frac{\pi}{4} + \frac{ax}{2}\right) - \frac{1}{a\sin ax}$
$\displaystyle \int \frac{dx}{\sin ax\cos^2 ax} = \frac{1}{a}\ln\tan\frac{ax}{2} + \frac{1}{a\cos ax}$
$\displaystyle \int \frac{dx}{\sin^2 ax\cos^2 ax} = -\frac{2\cot 2ax}{a}$
$\displaystyle \int \frac{\sin^2 ax}{\cos ax}\,dx = -\frac{\sin ax}{a} + \frac{1}{a}\ln\tan\left(\frac{ax}{2} + \frac{\pi}{4}\right)$
$\displaystyle \int \frac{\cos^2 ax}{\sin ax}\,dx = \frac{\cos ax}{a} + \frac{1}{a}\ln\tan\frac{ax}{2}$
$\displaystyle \int \frac{dx}{\cos ax(1 \pm \sin ax)} = \mp\frac{1}{2a(1 \pm \sin ax)} + \frac{1}{2a}\ln\tan\left(\frac{ax}{2} + \frac{\pi}{4}\right)$
$\displaystyle \int \frac{dx}{\sin ax(1 \pm \cos ax)} = \pm\frac{1}{2a(1 \pm \cos ax)} + \frac{1}{2a}\ln\tan\frac{ax}{2}$
$\displaystyle \int \frac{dx}{\sin ax \pm \cos ax} = \frac{1}{a\sqrt{2}}\ln\tan\left(\frac{ax}{2} \pm \frac{\pi}{8}\right)$
$\displaystyle \int \frac{\sin ax\,dx}{\sin ax \pm \cos ax} = \frac{x}{2} \mp \frac{1}{2a}\ln(\sin ax \pm\nobreak \cos ax)$
$\displaystyle \int \frac{\cos ax\,dx}{\sin ax \pm \cos ax} = \pm\frac{x}{2} + \frac{1}{2a}\ln(\sin ax \pm\nobreak \cos ax)$
$\displaystyle \int \frac{\sin ax\,dx}{p + q\cos ax} = -\frac{1}{aq}\ln(p +\nobreak q\cos ax)$
$\displaystyle \int \frac{\cos ax\,dx}{p + q\sin ax} = \frac{1}{aq}\ln(p +\nobreak q\sin ax)$
$\displaystyle \int \frac{\sin ax\,dx}{(p + q\cos ax)^n} = \frac{1}{aq(n - 1)(p + q\cos ax)^{n-1}}$
$\displaystyle \int \frac{\cos ax\,dx}{(p + q\sin ax)^n} = \frac{-1}{aq(n - 1)(p + q\sin ax)^{n-1}}$
$\displaystyle \int \frac{dx}{p\sin ax + q\cos ax} = \frac{1}{a\sqrt{p^2 + q^2}}\,\allowbreak\ln\tan\left(\frac{ax + \tan^{-1}(q/p)}{2}\right)$
$\displaystyle \int \frac{dx}{p\sin ax + q\cos ax + r} = \frac{2}{a\sqrt{r^2 - p^2 - q^2}}\,\allowbreak\tan^{-1}\left(\frac{p + (r - q)\tan(ax/2)}{\sqrt{r^2 - p^2 - q^2}}\right)$ $[r^2 > p^2 + q^2]$
$\displaystyle \int \frac{dx}{p\sin ax + q\cos ax + r} = \frac{1}{a\sqrt{p^2 + q^2 - r^2}}\,\allowbreak\ln\left(\frac{p - \sqrt{p^2 + q^2 - r^2} + (r - q)\tan(ax/2)}{p + \sqrt{p^2 + q^2 - r^2} + (r - q)\tan(ax/2)}\right)$ $[r^2 \lt p^2 + q^2]$
При $r = q$ вж. $\int \frac{dx}{p\sin ax + q(1 + \cos ax)}$. При $r^2 = p^2 + q^2$ вж. $\int \frac{dx}{p\sin ax + q\cos ax \pm \sqrt{p^2 + q^2}}$.
$\displaystyle \int \frac{dx}{p\sin ax + q(1 + \cos ax)} = \frac{1}{ap}\ln\left(q + p\tan\frac{ax}{2}\right)$
$\displaystyle \int \frac{dx}{p\sin ax + q\cos ax \pm \sqrt{p^2 + q^2}} = \frac{-1}{a\sqrt{p^2 + q^2}}\,\allowbreak\tan\left(\frac{\pi}{4} \mp \frac{ax + \tan^{-1}(q/p)}{2}\right)$
$\displaystyle \int \frac{dx}{p^2\sin^2 ax + q^2\cos^2 ax} = \frac{1}{apq}\tan^{-1}\left(\frac{p\tan ax}{q}\right)$
$\displaystyle \int \frac{dx}{p^2\sin^2 ax - q^2\cos^2 ax} = \frac{1}{2apq}\ln\left(\frac{p\tan ax - q}{p\tan ax + q}\right)$
$\displaystyle \int \sin^m ax\cos^n ax\,dx = -\frac{\sin^{m-1} ax\cos^{n+1} ax}{a(m + n)} + \frac{m - 1}{m + n}\int \sin^{m-2} ax\cos^n ax\,dx$
$\displaystyle = \frac{\sin^{m+1} ax\cos^{n-1} ax}{a(m + n)} + \frac{n - 1}{m + n}\int \sin^m ax\cos^{n-2} ax\,dx$
$\displaystyle \int \frac{\sin^m ax}{\cos^n ax}\,dx = \frac{\sin^{m-1} ax}{a(n - 1)\cos^{n-1} ax} - \frac{m - 1}{n - 1}\int \frac{\sin^{m-2} ax}{\cos^{n-2} ax}\,dx$
$\displaystyle = \frac{\sin^{m+1} ax}{a(n - 1)\cos^{n-1} ax} - \frac{m - n + 2}{n - 1}\int \frac{\sin^m ax}{\cos^{n-2} ax}\,dx$
$\displaystyle = \frac{-\sin^{m-1} ax}{a(m - n)\cos^{n-1} ax} + \frac{m - 1}{m - n}\int \frac{\sin^{m-2} ax}{\cos^n ax}\,dx$
$\displaystyle \int \frac{\cos^m ax}{\sin^n ax}\,dx = \frac{-\cos^{m-1} ax}{a(n - 1)\sin^{n-1} ax} - \frac{m - 1}{n - 1}\int \frac{\cos^{m-2} ax}{\sin^{n-2} ax}\,dx$
$\displaystyle = \frac{-\cos^{m+1} ax}{a(n - 1)\sin^{n-1} ax} - \frac{m - n + 2}{n - 1}\int \frac{\cos^m ax}{\sin^{n-2} ax}\,dx$
$\displaystyle = \frac{\cos^{m-1} ax}{a(m - n)\sin^{n-1} ax} + \frac{m - 1}{m - n}\int \frac{\cos^{m-2} ax}{\sin^n ax}\,dx$
$\displaystyle \int \frac{dx}{\sin^m ax\cos^n ax} = \frac{1}{a(n - 1)\sin^{m-1} ax\cos^{n-1} ax} + \frac{m + n - 2}{n - 1}\int \frac{dx}{\sin^m ax\cos^{n-2} ax}$
$\displaystyle = \frac{-1}{a(m - 1)\sin^{m-1} ax\cos^{n-1} ax} + \frac{m + n - 2}{m - 1}\,\allowbreak\int \frac{dx}{\sin^{m-2} ax\cos^n ax}$
Интеграли, съдържащи $\tan ax$
$\displaystyle \int \tan ax\,dx = -\frac{1}{a}\ln\cos ax = \frac{1}{a}\ln\sec ax$
$\displaystyle \int \tan^2 ax\,dx = \frac{\tan ax}{a} - x$
$\displaystyle \int \tan^3 ax\,dx = \frac{\tan^2 ax}{2a} + \frac{1}{a}\ln\cos ax$
$\displaystyle \int \tan^n ax\sec^2 ax\,dx = \frac{\tan^{n+1} ax}{(n + 1)a}$
$\displaystyle \int \frac{\sec^2 ax}{\tan ax}\,dx = \frac{1}{a}\ln\tan ax$
$\displaystyle \int \frac{dx}{\tan ax} = \frac{1}{a}\ln\sin ax$
$\displaystyle \int x\tan ax\,dx = \frac{1}{a^2}\bigg\{\frac{(ax)^3}{3} + \frac{(ax)^5}{15} + \frac{2(ax)^7}{105} + \cdots + \frac{2^{2n}(2^{2n} - 1)B_n(ax)^{2n+1}}{(2n + 1)!} + \cdots\bigg\}$
$\displaystyle \int \frac{\tan ax}{x}\,dx = ax + \frac{(ax)^3}{9} + \frac{2(ax)^5}{75} + \cdots + \frac{2^{2n}(2^{2n} - 1)B_n(ax)^{2n-1}}{(2n - 1)(2n)!} + \cdots$
$\displaystyle \int x\tan^2 ax\,dx = \frac{x\tan ax}{a} + \frac{1}{a^2}\ln\cos ax - \frac{x^2}{2}$
$\displaystyle \int \frac{dx}{p + q\tan ax} = \frac{px}{p^2 + q^2} + \frac{q}{a(p^2 + q^2)}\ln(q\sin ax +\nobreak p\cos ax)$
$\displaystyle \int \tan^n ax\,dx = \frac{\tan^{n-1} ax}{(n - 1)a} - \int \tan^{n-2} ax\,dx$
