МенюНеопределени Интеграли
Определение за неопределен интеграл
Ако $\frac{dy}{dx} = f(x)$, то $y$ е функция, чиято производна е $f(x)$, и се нарича примитивна функция на $f(x)$ или неопределен интеграл на $f(x)$ и се бележи с $\int f(x)\,dx$. Ако $y = \int f(u)\,du$, то $\frac{dy}{du} = f(u)$. Тъй като производната на константа е нула, всички неопределени интеграли на една функция се различават с константа.
Процесът на намиране на интеграл се нарича интегриране.
Основни правила на интегрирането
$u, v, w$ са функции на $x$;
$a, b, p, q, n$ са произволни константи (с ограничения, ако са посочени);
$e = 2{,}71828\ldots$ е основата на натуралните логаритми;
$\ln u$ е натуралният логаритъм от $u$, като $u > 0$ (по принцип, за да разширите формулите и за случая $u \lt 0$, заменете $\ln u$ с $\ln|u|$);
всички ъгли са в радиани;
всички интеграционни константи са пропуснати, но се подразбират.
$\displaystyle \int a\,dx = ax$
$\displaystyle \int af(x)\,dx = a\int f(x)\,dx$
$\displaystyle \int (u \pm\nobreak v \pm\nobreak w \pm\nobreak \cdots)\,dx = \int u\,dx \pm \int v\,dx \pm \int w\,dx \pm \cdots$
$\displaystyle \int u\,dv = uv - \int v\,du$ [интегриране по части]
$\displaystyle \int f(ax)\,dx = \frac{1}{a}\int f(u)\,du$
$\displaystyle \int F[f(x)]\,dx = \int F(u)\frac{dx}{du}\,du = \int \frac{F(u)}{f'(x)}\,du$ [където $u = f(x)$]
$\displaystyle \int u^n\,du = \frac{u^{n+1}}{n + 1}$ $[n \ne -1]$
$\displaystyle \int \frac{du}{u} = \ln|u|$ [$\ln u$ при $u > 0$, $\ln(-u)$ при $u \lt 0$]
$\displaystyle \int e^u\,du = e^u$
$\displaystyle \int a^u\,du = \int e^{u\ln a}\,du = \frac{e^{u\ln a}}{\ln a} = \frac{a^u}{\ln a}$ $[a > 0,\ a \ne 1]$
$\displaystyle \int \sin u\,du = -\cos u$
$\displaystyle \int \cos u\,du = \sin u$
$\displaystyle \int \tan u\,du = \ln\sec u = -\ln\cos u$
$\displaystyle \int \cot u\,du = \ln\sin u$
$\displaystyle \int \sec u\,du = \ln(\sec u +\nobreak \tan u) = \ln\tan\left(\frac{u}{2} + \frac{\pi}{4}\right)$
$\displaystyle \int \csc u\,du = \ln(\csc u -\nobreak \cot u) = \ln\tan\frac{u}{2}$
$\displaystyle \int \sec^2 u\,du = \tan u$
$\displaystyle \int \csc^2 u\,du = -\cot u$
$\displaystyle \int \tan^2 u\,du = \tan u - u$
$\displaystyle \int \cot^2 u\,du = -\cot u - u$
$\displaystyle \int \sin^2 u\,du = \frac{u}{2} - \frac{\sin 2u}{4} = \frac{1}{2}(u -\nobreak \sin u\cos u)$
$\displaystyle \int \cos^2 u\,du = \frac{u}{2} + \frac{\sin 2u}{4} = \frac{1}{2}(u +\nobreak \sin u\cos u)$
$\displaystyle \int \sec u\tan u\,du = \sec u$
$\displaystyle \int \csc u\cot u\,du = -\csc u$
$\displaystyle \int \sinh u\,du = \cosh u$
$\displaystyle \int \cosh u\,du = \sinh u$
$\displaystyle \int \tanh u\,du = \ln\cosh u$
$\displaystyle \int \coth u\,du = \ln\sinh u$
$\displaystyle \int \operatorname{sech} u\,du = \sin^{-1}(\tanh u)\allowbreak\quad\text{или}\quad 2\tan^{-1}e^u$
$\displaystyle \int \operatorname{csch} u\,du = \ln\tanh\frac{u}{2}\allowbreak\quad\text{или}\quad -2\coth^{-1}e^u$
$\displaystyle \int \operatorname{sech}^2 u\,du = \tanh u$
$\displaystyle \int \operatorname{csch}^2 u\,du = -\coth u$
$\displaystyle \int \tanh^2 u\,du = u - \tanh u$
$\displaystyle \int \coth^2 u\,du = u - \coth u$
$\displaystyle \int \sinh^2 u\,du = \frac{\sinh 2u}{4} - \frac{u}{2} = \frac{1}{2}(\sinh u\cosh u -\nobreak u)$
$\displaystyle \int \cosh^2 u\,du = \frac{\sinh 2u}{4} + \frac{u}{2} = \frac{1}{2}(\sinh u\cosh u +\nobreak u)$
$\displaystyle \int \operatorname{sech} u\tanh u\,du = -\operatorname{sech} u$
$\displaystyle \int \operatorname{csch} u\coth u\,du = -\operatorname{csch} u$
$\displaystyle \int \frac{du}{u^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{u}{a}$
$\displaystyle \int \frac{du}{u^2 - a^2} = \frac{1}{2a}\ln\left(\frac{u - a}{u + a}\right) = -\frac{1}{a}\coth^{-1}\frac{u}{a}$ $[u^2 > a^2]$
$\displaystyle \int \frac{du}{a^2 - u^2} = \frac{1}{2a}\ln\left(\frac{a + u}{a - u}\right) = \frac{1}{a}\tanh^{-1}\frac{u}{a}$ $[u^2 \lt a^2]$
$\displaystyle \int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a}$
$\displaystyle \int \frac{du}{\sqrt{u^2 + a^2}} = \ln\left(u + \sqrt{u^2 + a^2}\right)\allowbreak\quad\text{или}\quad\sinh^{-1}\frac{u}{a}$
$\displaystyle \int \frac{du}{\sqrt{u^2 - a^2}} = \ln\left(u + \sqrt{u^2 - a^2}\right)$
$\displaystyle \int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a}\sec^{-1}\left|\frac{u}{a}\right|$
$\displaystyle \int \frac{du}{u\sqrt{u^2 + a^2}} = -\frac{1}{a}\ln\left(\frac{a + \sqrt{u^2 + a^2}}{u}\right)$
$\displaystyle \int \frac{du}{u\sqrt{a^2 - u^2}} = -\frac{1}{a}\ln\left(\frac{a + \sqrt{a^2 - u^2}}{u}\right)$
$\displaystyle \int f^{(n)}g\,dx = f^{(n-1)}g - f^{(n-2)}g' + f^{(n-3)}g'' - \cdots + (-1)^n\int fg^{(n)}\,dx$
Това се нарича обобщено интегриране по части.
Важни трансформации
Често на практика интегралът може да бъде опростен чрез подходяща трансформация или субституция и правилото за смяна на променливата (формулата за $\int F[f(x)]\,dx$ по-горе). Следният списък показва някои трансформации и резултата от тях.
$\displaystyle \int F(ax +\nobreak b)\,dx = \frac{1}{a}\int F(u)\,du$ [където $u = ax + b$]
$\displaystyle \int F\left(\sqrt{ax + b}\right)dx = \frac{2}{a}\int uF(u)\,du$ [където $u = \sqrt{ax + b}$]
$\displaystyle \int F\left(\sqrt[n]{ax + b}\right)dx = \frac{n}{a}\int u^{n-1}F(u)\,du$ [където $u = \sqrt[n]{ax + b}$]
$\displaystyle \int F\left(\sqrt{a^2 - x^2}\right)dx = a\int F(a\cos u)\cos u\,du$ [където $x = a\sin u$]
$\displaystyle \int F\left(\sqrt{x^2 + a^2}\right)dx = a\int F(a\sec u)\sec^2 u\,du$ [където $x = a\tan u$]
$\displaystyle \int F\left(\sqrt{x^2 - a^2}\right)dx = a\int F(a\tan u)\sec u\tan u\,du$ [където $x = a\sec u$]
$\displaystyle \int F(e^{ax})\,dx = \frac{1}{a}\int \frac{F(u)}{u}\,du$ [където $u = e^{ax}$]
$\displaystyle \int F(\ln x)\,dx = \int F(u)e^u\,du$ [където $u = \ln x$]
$\displaystyle \int F\left(\sin^{-1}\frac{x}{a}\right)dx = a\int F(u)\cos u\,du$ [където $u = \sin^{-1}\frac{x}{a}$]
Подобни резултати са в сила и за другите обратни тригонометрични функции.
$\displaystyle \int F(\sin x, \cos x)\,dx = 2\int F\left(\frac{2u}{1 + u^2}, \frac{1 - u^2}{1 + u^2}\right)\frac{du}{1 + u^2}$ [където $u = \tan\frac{x}{2}$]
Интеграли, съдържащи $ax + b$
$\displaystyle \int \frac{dx}{ax + b} = \frac{1}{a}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x\,dx}{ax + b} = \frac{x}{a} - \frac{b}{a^2}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x^2\,dx}{ax + b} = \frac{(ax + b)^2}{2a^3} - \frac{2b(ax + b)}{a^3} + \frac{b^2}{a^3}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x^3\,dx}{ax + b} = \frac{(ax + b)^3}{3a^4} - \frac{3b(ax + b)^2}{2a^4} + \frac{3b^2(ax + b)}{a^4} - \frac{b^3}{a^4}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{dx}{x(ax + b)} = \frac{1}{b}\ln\left(\frac{x}{ax + b}\right)$
$\displaystyle \int \frac{dx}{x^2(ax + b)} = -\frac{1}{bx} + \frac{a}{b^2}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int \frac{dx}{x^3(ax + b)} = \frac{2ax - b}{2b^2x^2} + \frac{a^2}{b^3}\ln\left(\frac{x}{ax + b}\right)$
$\displaystyle \int \frac{dx}{(ax + b)^2} = \frac{-1}{a(ax + b)}$
$\displaystyle \int \frac{x\,dx}{(ax + b)^2} = \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x^2\,dx}{(ax + b)^2} = \frac{ax + b}{a^3} - \frac{b^2}{a^3(ax + b)} - \frac{2b}{a^3}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x^3\,dx}{(ax + b)^2} = \frac{(ax + b)^2}{2a^4} - \frac{3b(ax + b)}{a^4} + \frac{b^3}{a^4(ax + b)} + \frac{3b^2}{a^4}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{dx}{x(ax + b)^2} = \frac{1}{b(ax + b)} + \frac{1}{b^2}\ln\left(\frac{x}{ax + b}\right)$
$\displaystyle \int \frac{dx}{x^2(ax + b)^2} = \frac{-a}{b^2(ax + b)} - \frac{1}{b^2x} + \frac{2a}{b^3}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int \frac{dx}{x^3(ax + b)^2} = -\frac{(ax + b)^2}{2b^4x^2} + \frac{3a(ax + b)}{b^4x} - \frac{a^3x}{b^4(ax + b)} - \frac{3a^2}{b^4}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int \frac{dx}{(ax + b)^3} = \frac{-1}{2a(ax + b)^2}$
$\displaystyle \int \frac{x\,dx}{(ax + b)^3} = \frac{-1}{a^2(ax + b)} + \frac{b}{2a^2(ax + b)^2}$
$\displaystyle \int \frac{x^2\,dx}{(ax + b)^3} = \frac{2b}{a^3(ax + b)} - \frac{b^2}{2a^3(ax + b)^2} + \frac{1}{a^3}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{x^3\,dx}{(ax + b)^3} = \frac{x}{a^3} - \frac{3b^2}{a^4(ax + b)} + \frac{b^3}{2a^4(ax + b)^2} - \frac{3b}{a^4}\ln(ax +\nobreak b)$
$\displaystyle \int \frac{dx}{x(ax + b)^3} = \frac{a^2x^2}{2b^3(ax + b)^2} - \frac{2ax}{b^3(ax + b)} - \frac{1}{b^3}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int \frac{dx}{x^2(ax + b)^3} = \frac{-a}{2b^2(ax + b)^2} - \frac{2a}{b^3(ax + b)} - \frac{1}{b^3x} + \frac{3a}{b^4}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int \frac{dx}{x^3(ax + b)^3} = \frac{a^4x^2}{2b^5(ax + b)^2} - \frac{4a^3x}{b^5(ax + b)} - \frac{(ax + b)^2}{2b^5x^2} + \frac{4a(ax + b)}{b^5x} - \frac{6a^2}{b^5}\ln\left(\frac{ax + b}{x}\right)$
$\displaystyle \int (ax +\nobreak b)^n\,dx = \frac{(ax + b)^{n+1}}{(n + 1)a}$ [при $n = -1$, вж. $\int \frac{dx}{ax + b}$]
$\displaystyle \int x(ax +\nobreak b)^n\,dx = \frac{(ax + b)^{n+2}}{(n + 2)a^2} - \frac{b(ax + b)^{n+1}}{(n + 1)a^2}$ [при $n = -1, -2$, вж. $\int \frac{x\,dx}{ax + b}$, $\int \frac{x\,dx}{(ax + b)^2}$]
$\displaystyle \int x^2(ax +\nobreak b)^n\,dx = \frac{(ax + b)^{n+3}}{(n + 3)a^3} - \frac{2b(ax + b)^{n+2}}{(n + 2)a^3} + \frac{b^2(ax + b)^{n+1}}{(n + 1)a^3}$ [при $n = -1, -2, -3$, вж. $\int \frac{x^2\,dx}{ax + b}$, $\int \frac{x^2\,dx}{(ax + b)^2}$, $\int \frac{x^2\,dx}{(ax + b)^3}$]
$\displaystyle \int x^m(ax +\nobreak b)^n\,dx = \frac{x^{m+1}(ax + b)^n}{m + n + 1} + \frac{nb}{m + n + 1}\,\allowbreak\int x^m(ax +\nobreak b)^{n-1}\,dx$
$\displaystyle = \frac{x^m(ax + b)^{n+1}}{(m + n + 1)a} - \frac{mb}{(m + n + 1)a}\,\allowbreak\int x^{m-1}(ax +\nobreak b)^n\,dx$
$\displaystyle = \frac{-x^{m+1}(ax + b)^{n+1}}{(n + 1)b} + \frac{m + n + 2}{(n + 1)b}\,\allowbreak\int x^m(ax +\nobreak b)^{n+1}\,dx$
Интеграли, съдържащи $\sqrt{ax + b}$
$\displaystyle \int \frac{dx}{\sqrt{ax + b}} = \frac{2\sqrt{ax + b}}{a}$
$\displaystyle \int \frac{x\,dx}{\sqrt{ax + b}} = \frac{2(ax - 2b)}{3a^2}\sqrt{ax + b}$
$\displaystyle \int \frac{x^2\,dx}{\sqrt{ax + b}} = \frac{2(3a^2x^2 - 4abx + 8b^2)}{15a^3}\sqrt{ax + b}$
$\displaystyle \int \frac{dx}{x\sqrt{ax + b}} = \frac{1}{\sqrt{b}}\,\allowbreak\ln\left(\frac{\sqrt{ax + b} - \sqrt{b}}{\sqrt{ax + b} + \sqrt{b}}\right)$ $[b > 0]$
$\displaystyle \int \frac{dx}{x\sqrt{ax + b}} = \frac{2}{\sqrt{-b}}\tan^{-1}\sqrt{\frac{ax + b}{-b}}$ $[b \lt 0]$
$\displaystyle \int \frac{dx}{x^2\sqrt{ax + b}} = -\frac{\sqrt{ax + b}}{bx} - \frac{a}{2b}\int \frac{dx}{x\sqrt{ax + b}}$
$\displaystyle \int \sqrt{ax + b}\,dx = \frac{2\sqrt{(ax + b)^3}}{3a}$
$\displaystyle \int x\sqrt{ax + b}\,dx = \frac{2(3ax - 2b)}{15a^2}\sqrt{(ax + b)^3}$
$\displaystyle \int x^2\sqrt{ax + b}\,dx = \frac{2(15a^2x^2 - 12abx + 8b^2)}{105a^3}\,\allowbreak\sqrt{(ax + b)^3}$
$\displaystyle \int \frac{\sqrt{ax + b}}{x}\,dx = 2\sqrt{ax + b} + b\int \frac{dx}{x\sqrt{ax + b}}$
$\displaystyle \int \frac{\sqrt{ax + b}}{x^2}\,dx = -\frac{\sqrt{ax + b}}{x} + \frac{a}{2}\int \frac{dx}{x\sqrt{ax + b}}$
$\displaystyle \int \frac{x^m}{\sqrt{ax + b}}\,dx = \frac{2x^m\sqrt{ax + b}}{(2m + 1)a} - \frac{2mb}{(2m + 1)a}\,\allowbreak\int \frac{x^{m-1}}{\sqrt{ax + b}}\,dx$
$\displaystyle \int \frac{dx}{x^m\sqrt{ax + b}} = -\frac{\sqrt{ax + b}}{(m - 1)bx^{m-1}} - \frac{(2m - 3)a}{(2m - 2)b}\,\allowbreak\int \frac{dx}{x^{m-1}\sqrt{ax + b}}$
$\displaystyle \int x^m\sqrt{ax + b}\,dx = \frac{2x^m}{(2m + 3)a}(ax +\nobreak b)^{3/2} - \frac{2mb}{(2m + 3)a}\,\allowbreak\int x^{m-1}\sqrt{ax + b}\,dx$
$\displaystyle \int \frac{\sqrt{ax + b}}{x^m}\,dx = -\frac{\sqrt{ax + b}}{(m - 1)x^{m-1}} + \frac{a}{2(m - 1)}\,\allowbreak\int \frac{dx}{x^{m-1}\sqrt{ax + b}}$
$\displaystyle \int \frac{\sqrt{ax + b}}{x^m}\,dx = \frac{-(ax + b)^{3/2}}{(m - 1)bx^{m-1}} - \frac{(2m - 5)a}{(2m - 2)b}\,\allowbreak\int \frac{\sqrt{ax + b}}{x^{m-1}}\,dx$
$\displaystyle \int (ax +\nobreak b)^{m/2}\,dx = \frac{2(ax + b)^{(m+2)/2}}{a(m + 2)}$
$\displaystyle \int x(ax +\nobreak b)^{m/2}\,dx = \frac{2(ax + b)^{(m+4)/2}}{a^2(m + 4)} - \frac{2b(ax + b)^{(m+2)/2}}{a^2(m + 2)}$
$\displaystyle \int x^2(ax +\nobreak b)^{m/2}\,dx = \frac{2(ax + b)^{(m+6)/2}}{a^3(m + 6)} - \frac{4b(ax + b)^{(m+4)/2}}{a^3(m + 4)} + \frac{2b^2(ax + b)^{(m+2)/2}}{a^3(m + 2)}$
$\displaystyle \int \frac{(ax + b)^{m/2}}{x}\,dx = \frac{2(ax + b)^{m/2}}{m} + b\int \frac{(ax + b)^{(m-2)/2}}{x}\,dx$
$\displaystyle \int \frac{(ax + b)^{m/2}}{x^2}\,dx = -\frac{(ax + b)^{(m+2)/2}}{bx} + \frac{ma}{2b}\int \frac{(ax + b)^{m/2}}{x}\,dx$
$\displaystyle \int \frac{dx}{x(ax + b)^{m/2}} = \frac{2}{(m - 2)b(ax + b)^{(m-2)/2}} + \frac{1}{b}\int \frac{dx}{x(ax + b)^{(m-2)/2}}$
Интеграли, съдържащи $ax + b$ и $px + q$
$\displaystyle \int \frac{dx}{(ax + b)(px + q)} = \frac{1}{bp - aq}\ln\left(\frac{px + q}{ax + b}\right)$
$\displaystyle \int \frac{x\,dx}{(ax + b)(px + q)} = \frac{1}{bp - aq}\,\allowbreak\bigg\{\frac{b}{a}\ln(ax +\nobreak b) - \frac{q}{p}\ln(px +\nobreak q)\bigg\}$
$\displaystyle \int \frac{dx}{(ax + b)^2(px + q)} = \frac{1}{bp - aq}\,\allowbreak\bigg\{\frac{1}{ax + b} + \frac{p}{bp - aq}\ln\left(\frac{px + q}{ax + b}\right)\bigg\}$
$\displaystyle \int \frac{x\,dx}{(ax + b)^2(px + q)} = \frac{1}{bp - aq}\,\allowbreak\bigg\{\frac{q}{bp - aq}\ln\left(\frac{ax + b}{px + q}\right) - \frac{b}{a(ax + b)}\bigg\}$
$\displaystyle \int \frac{x^2\,dx}{(ax + b)^2(px + q)} = \frac{b^2}{(bp - aq)a^2(ax + b)} + \frac{1}{(bp - aq)^2}\,\allowbreak\bigg\{\frac{q^2}{p}\ln(px +\nobreak q) + \frac{b(bp - 2aq)}{a^2}\ln(ax +\nobreak b)\bigg\}$
$\displaystyle \int \frac{dx}{(ax + b)^m(px + q)^n} = \frac{-1}{(n - 1)(bp - aq)}\,\allowbreak\bigg\{\frac{1}{(ax + b)^{m-1}(px + q)^{n-1}} + a(m +\nobreak n -\nobreak 2)\,\allowbreak\int \frac{dx}{(ax + b)^m(px + q)^{n-1}}\bigg\}$
$\displaystyle \int \frac{ax + b}{px + q}\,dx = \frac{ax}{p} + \frac{bp - aq}{p^2}\ln(px +\nobreak q)$
$\displaystyle \int \frac{(ax + b)^m}{(px + q)^n}\,dx = \frac{-1}{(n - 1)(bp - aq)}\bigg\{\frac{(ax + b)^{m+1}}{(px + q)^{n-1}} + (n -\nobreak m -\nobreak 2)a\,\allowbreak\int \frac{(ax + b)^m}{(px + q)^{n-1}}\,dx\bigg\}$
$\displaystyle = \frac{-1}{(n - m - 1)p}\bigg\{\frac{(ax + b)^m}{(px + q)^{n-1}} + m(bp -\nobreak aq)\,\allowbreak\int \frac{(ax + b)^{m-1}}{(px + q)^n}\,dx\bigg\}$
$\displaystyle = \frac{-1}{(n - 1)p}\bigg\{\frac{(ax + b)^m}{(px + q)^{n-1}} - ma\,\allowbreak\int \frac{(ax + b)^{m-1}}{(px + q)^{n-1}}\,dx\bigg\}$
Интеграли, съдържащи $\sqrt{ax + b}$ и $px + q$
$\displaystyle \int \frac{px + q}{\sqrt{ax + b}}\,dx = \frac{2(apx + 3aq - 2bp)}{3a^2}\sqrt{ax + b}$
$\displaystyle \int \frac{dx}{(px + q)\sqrt{ax + b}} = \frac{1}{\sqrt{bp - aq}\sqrt{p}}\,\allowbreak\ln\left(\frac{\sqrt{p(ax + b)} - \sqrt{bp - aq}}{\sqrt{p(ax + b)} + \sqrt{bp - aq}}\right)$ $[bp > aq]$
$\displaystyle \int \frac{dx}{(px + q)\sqrt{ax + b}} = \frac{2}{\sqrt{aq - bp}\sqrt{p}}\,\allowbreak\tan^{-1}\sqrt{\frac{p(ax + b)}{aq - bp}}$ $[bp \lt aq]$
$\displaystyle \int \frac{\sqrt{ax + b}}{px + q}\,dx = \frac{2\sqrt{ax + b}}{p} + \frac{\sqrt{bp - aq}}{p\sqrt{p}}\,\allowbreak\ln\left(\frac{\sqrt{p(ax + b)} - \sqrt{bp - aq}}{\sqrt{p(ax + b)} + \sqrt{bp - aq}}\right)$ $[bp > aq]$
$\displaystyle \int \frac{\sqrt{ax + b}}{px + q}\,dx = \frac{2\sqrt{ax + b}}{p} - \frac{2\sqrt{aq - bp}}{p\sqrt{p}}\,\allowbreak\tan^{-1}\sqrt{\frac{p(ax + b)}{aq - bp}}$ $[bp \lt aq]$
$\displaystyle \int (px +\nobreak q)^n\sqrt{ax + b}\,dx = \frac{2(px + q)^{n+1}\sqrt{ax + b}}{(2n + 3)p} + \frac{bp - aq}{(2n + 3)p}\,\allowbreak\int \frac{(px + q)^n}{\sqrt{ax + b}}\,dx$
$\displaystyle \int \frac{dx}{(px + q)^n\sqrt{ax + b}} = \frac{\sqrt{ax + b}}{(n - 1)(aq - bp)(px + q)^{n-1}} + \frac{(2n - 3)a}{2(n - 1)(aq - bp)}\,\allowbreak\int \frac{dx}{(px + q)^{n-1}\sqrt{ax + b}}$
$\displaystyle \int \frac{(px + q)^n}{\sqrt{ax + b}}\,dx = \frac{2(px + q)^n\sqrt{ax + b}}{(2n + 1)a} + \frac{2n(aq - bp)}{(2n + 1)a}\,\allowbreak\int \frac{(px + q)^{n-1}\,dx}{\sqrt{ax + b}}$
$\displaystyle \int \frac{\sqrt{ax + b}}{(px + q)^n}\,dx = \frac{-\sqrt{ax + b}}{(n - 1)p(px + q)^{n-1}} + \frac{a}{2(n - 1)p}\,\allowbreak\int \frac{dx}{(px + q)^{n-1}\sqrt{ax + b}}$
Интеграли, съдържащи $\sqrt{ax + b}$ и $\sqrt{px + q}$
$\displaystyle \int \frac{dx}{\sqrt{(ax + b)(px + q)}} = \frac{2}{\sqrt{ap}}\,\allowbreak\ln\left(\sqrt{a(px + q)} + \sqrt{p(ax + b)}\right)$ $[ap > 0]$
$\displaystyle \int \frac{dx}{\sqrt{(ax + b)(px + q)}} = \frac{2}{\sqrt{-ap}}\tan^{-1}\sqrt{\frac{-p(ax + b)}{a(px + q)}}$ $[ap \lt 0]$
$\displaystyle \int \frac{x\,dx}{\sqrt{(ax + b)(px + q)}} = \frac{\sqrt{(ax + b)(px + q)}}{ap} - \frac{bp + aq}{2ap}\int \frac{dx}{\sqrt{(ax + b)(px + q)}}$
$\displaystyle \int \sqrt{(ax + b)(px + q)}\,dx = \frac{2apx + bp + aq}{4ap}\,\allowbreak\sqrt{(ax + b)(px + q)} - \frac{(bp - aq)^2}{8ap}\int \frac{dx}{\sqrt{(ax + b)(px + q)}}$
$\displaystyle \int \sqrt{\frac{px + q}{ax + b}}\,dx = \frac{\sqrt{(ax + b)(px + q)}}{a} + \frac{aq - bp}{2a}\int \frac{dx}{\sqrt{(ax + b)(px + q)}}$
$\displaystyle \int \frac{dx}{(px + q)\sqrt{(ax + b)(px + q)}} = \frac{2\sqrt{ax + b}}{(aq - bp)\sqrt{px + q}}$
Интеграли, съдържащи $x^2 + a^2$
$\displaystyle \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{x^2 + a^2} = \frac{1}{2}\ln(x^2 +\nobreak a^2)$
$\displaystyle \int \frac{x^2\,dx}{x^2 + a^2} = x - a\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x^3\,dx}{x^2 + a^2} = \frac{x^2}{2} - \frac{a^2}{2}\ln(x^2 +\nobreak a^2)$
$\displaystyle \int \frac{dx}{x(x^2 + a^2)} = \frac{1}{2a^2}\ln\left(\frac{x^2}{x^2 + a^2}\right)$
$\displaystyle \int \frac{dx}{x^2(x^2 + a^2)} = -\frac{1}{a^2x} - \frac{1}{a^3}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{dx}{x^3(x^2 + a^2)} = -\frac{1}{2a^2x^2} - \frac{1}{2a^4}\ln\left(\frac{x^2}{x^2 + a^2}\right)$
$\displaystyle \int \frac{dx}{(x^2 + a^2)^2} = \frac{x}{2a^2(x^2 + a^2)} + \frac{1}{2a^3}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{(x^2 + a^2)^2} = \frac{-1}{2(x^2 + a^2)}$
$\displaystyle \int \frac{x^2\,dx}{(x^2 + a^2)^2} = \frac{-x}{2(x^2 + a^2)} + \frac{1}{2a}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x^3\,dx}{(x^2 + a^2)^2} = \frac{a^2}{2(x^2 + a^2)} + \frac{1}{2}\ln(x^2 +\nobreak a^2)$
$\displaystyle \int \frac{dx}{x(x^2 + a^2)^2} = \frac{1}{2a^2(x^2 + a^2)} + \frac{1}{2a^4}\ln\left(\frac{x^2}{x^2 + a^2}\right)$
$\displaystyle \int \frac{dx}{x^2(x^2 + a^2)^2} = -\frac{1}{a^4x} - \frac{x}{2a^4(x^2 + a^2)} - \frac{3}{2a^5}\tan^{-1}\frac{x}{a}$
$\displaystyle \int \frac{dx}{x^3(x^2 + a^2)^2} = -\frac{1}{2a^4x^2} - \frac{1}{2a^4(x^2 + a^2)} - \frac{1}{a^6}\ln\left(\frac{x^2}{x^2 + a^2}\right)$
$\displaystyle \int \frac{dx}{(x^2 + a^2)^n} = \frac{x}{2(n - 1)a^2(x^2 + a^2)^{n-1}} + \frac{2n - 3}{(2n - 2)a^2}\int \frac{dx}{(x^2 + a^2)^{n-1}}$
$\displaystyle \int \frac{x\,dx}{(x^2 + a^2)^n} = \frac{-1}{2(n - 1)(x^2 + a^2)^{n-1}}$
$\displaystyle \int \frac{dx}{x(x^2 + a^2)^n} = \frac{1}{2(n - 1)a^2(x^2 + a^2)^{n-1}} + \frac{1}{a^2}\int \frac{dx}{x(x^2 + a^2)^{n-1}}$
$\displaystyle \int \frac{x^m\,dx}{(x^2 + a^2)^n} = \int \frac{x^{m-2}\,dx}{(x^2 + a^2)^{n-1}} - a^2\int \frac{x^{m-2}\,dx}{(x^2 + a^2)^n}$
$\displaystyle \int \frac{dx}{x^m(x^2 + a^2)^n} = \frac{1}{a^2}\int \frac{dx}{x^m(x^2 + a^2)^{n-1}} - \frac{1}{a^2}\int \frac{dx}{x^{m-2}(x^2 + a^2)^n}$
Интеграли, съдържащи $x^2 - a^2$, $x^2 > a^2$
$\displaystyle \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ln\left(\frac{x - a}{x + a}\right)\allowbreak\quad\text{или}\quad -\frac{1}{a}\coth^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{x^2 - a^2} = \frac{1}{2}\ln(x^2 -\nobreak a^2)$
$\displaystyle \int \frac{x^2\,dx}{x^2 - a^2} = x + \frac{a}{2}\ln\left(\frac{x - a}{x + a}\right)$
$\displaystyle \int \frac{x^3\,dx}{x^2 - a^2} = \frac{x^2}{2} + \frac{a^2}{2}\ln(x^2 -\nobreak a^2)$
$\displaystyle \int \frac{dx}{x(x^2 - a^2)} = \frac{1}{2a^2}\ln\left(\frac{x^2 - a^2}{x^2}\right)$
$\displaystyle \int \frac{dx}{x^2(x^2 - a^2)} = \frac{1}{a^2x} + \frac{1}{2a^3}\ln\left(\frac{x - a}{x + a}\right)$
$\displaystyle \int \frac{dx}{x^3(x^2 - a^2)} = \frac{1}{2a^2x^2} - \frac{1}{2a^4}\ln\left(\frac{x^2}{x^2 - a^2}\right)$
$\displaystyle \int \frac{dx}{(x^2 - a^2)^2} = \frac{-x}{2a^2(x^2 - a^2)} - \frac{1}{4a^3}\ln\left(\frac{x - a}{x + a}\right)$
$\displaystyle \int \frac{x\,dx}{(x^2 - a^2)^2} = \frac{-1}{2(x^2 - a^2)}$
$\displaystyle \int \frac{x^2\,dx}{(x^2 - a^2)^2} = \frac{-x}{2(x^2 - a^2)} + \frac{1}{4a}\ln\left(\frac{x - a}{x + a}\right)$
$\displaystyle \int \frac{x^3\,dx}{(x^2 - a^2)^2} = \frac{-a^2}{2(x^2 - a^2)} + \frac{1}{2}\ln(x^2 -\nobreak a^2)$
$\displaystyle \int \frac{dx}{x(x^2 - a^2)^2} = \frac{-1}{2a^2(x^2 - a^2)} + \frac{1}{2a^4}\ln\left(\frac{x^2}{x^2 - a^2}\right)$
$\displaystyle \int \frac{dx}{x^2(x^2 - a^2)^2} = -\frac{1}{a^4x} - \frac{x}{2a^4(x^2 - a^2)} - \frac{3}{4a^5}\ln\left(\frac{x - a}{x + a}\right)$
$\displaystyle \int \frac{dx}{x^3(x^2 - a^2)^2} = -\frac{1}{2a^4x^2} - \frac{1}{2a^4(x^2 - a^2)} + \frac{1}{a^6}\ln\left(\frac{x^2}{x^2 - a^2}\right)$
$\displaystyle \int \frac{dx}{(x^2 - a^2)^n} = \frac{-x}{2(n - 1)a^2(x^2 - a^2)^{n-1}} - \frac{2n - 3}{(2n - 2)a^2}\int \frac{dx}{(x^2 - a^2)^{n-1}}$
$\displaystyle \int \frac{x\,dx}{(x^2 - a^2)^n} = \frac{-1}{2(n - 1)(x^2 - a^2)^{n-1}}$
$\displaystyle \int \frac{dx}{x(x^2 - a^2)^n} = \frac{-1}{2(n - 1)a^2(x^2 - a^2)^{n-1}} - \frac{1}{a^2}\int \frac{dx}{x(x^2 - a^2)^{n-1}}$
$\displaystyle \int \frac{x^m\,dx}{(x^2 - a^2)^n} = \int \frac{x^{m-2}\,dx}{(x^2 - a^2)^{n-1}} + a^2\int \frac{x^{m-2}\,dx}{(x^2 - a^2)^n}$
$\displaystyle \int \frac{dx}{x^m(x^2 - a^2)^n} = \frac{1}{a^2}\int \frac{dx}{x^{m-2}(x^2 - a^2)^n} - \frac{1}{a^2}\int \frac{dx}{x^m(x^2 - a^2)^{n-1}}$
Интеграли, съдържащи $a^2 - x^2$, $x^2 \lt a^2$
$\displaystyle \int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\left(\frac{a + x}{a - x}\right)\allowbreak\quad\text{или}\quad \frac{1}{a}\tanh^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{a^2 - x^2} = -\frac{1}{2}\ln(a^2 -\nobreak x^2)$
$\displaystyle \int \frac{x^2\,dx}{a^2 - x^2} = -x + \frac{a}{2}\ln\left(\frac{a + x}{a - x}\right)$
$\displaystyle \int \frac{x^3\,dx}{a^2 - x^2} = -\frac{x^2}{2} - \frac{a^2}{2}\ln(a^2 -\nobreak x^2)$
$\displaystyle \int \frac{dx}{x(a^2 - x^2)} = \frac{1}{2a^2}\ln\left(\frac{x^2}{a^2 - x^2}\right)$
$\displaystyle \int \frac{dx}{x^2(a^2 - x^2)} = -\frac{1}{a^2x} + \frac{1}{2a^3}\ln\left(\frac{a + x}{a - x}\right)$
$\displaystyle \int \frac{dx}{x^3(a^2 - x^2)} = -\frac{1}{2a^2x^2} + \frac{1}{2a^4}\ln\left(\frac{x^2}{a^2 - x^2}\right)$
$\displaystyle \int \frac{dx}{(a^2 - x^2)^2} = \frac{x}{2a^2(a^2 - x^2)} + \frac{1}{4a^3}\ln\left(\frac{a + x}{a - x}\right)$
$\displaystyle \int \frac{x\,dx}{(a^2 - x^2)^2} = \frac{1}{2(a^2 - x^2)}$
$\displaystyle \int \frac{x^2\,dx}{(a^2 - x^2)^2} = \frac{x}{2(a^2 - x^2)} - \frac{1}{4a}\ln\left(\frac{a + x}{a - x}\right)$
$\displaystyle \int \frac{x^3\,dx}{(a^2 - x^2)^2} = \frac{a^2}{2(a^2 - x^2)} + \frac{1}{2}\ln(a^2 -\nobreak x^2)$
$\displaystyle \int \frac{dx}{x(a^2 - x^2)^2} = \frac{1}{2a^2(a^2 - x^2)} + \frac{1}{2a^4}\ln\left(\frac{x^2}{a^2 - x^2}\right)$
$\displaystyle \int \frac{dx}{x^2(a^2 - x^2)^2} = \frac{-1}{a^4x} + \frac{x}{2a^4(a^2 - x^2)} + \frac{3}{4a^5}\ln\left(\frac{a + x}{a - x}\right)$
$\displaystyle \int \frac{dx}{x^3(a^2 - x^2)^2} = \frac{-1}{2a^4x^2} + \frac{1}{2a^4(a^2 - x^2)} + \frac{1}{a^6}\ln\left(\frac{x^2}{a^2 - x^2}\right)$
$\displaystyle \int \frac{dx}{(a^2 - x^2)^n} = \frac{x}{2(n - 1)a^2(a^2 - x^2)^{n-1}} + \frac{2n - 3}{(2n - 2)a^2}\int \frac{dx}{(a^2 - x^2)^{n-1}}$
$\displaystyle \int \frac{x\,dx}{(a^2 - x^2)^n} = \frac{1}{2(n - 1)(a^2 - x^2)^{n-1}}$
$\displaystyle \int \frac{dx}{x(a^2 - x^2)^n} = \frac{1}{2(n - 1)a^2(a^2 - x^2)^{n-1}} + \frac{1}{a^2}\int \frac{dx}{x(a^2 - x^2)^{n-1}}$
$\displaystyle \int \frac{x^m\,dx}{(a^2 - x^2)^n} = a^2\int \frac{x^{m-2}\,dx}{(a^2 - x^2)^n} - \int \frac{x^{m-2}\,dx}{(a^2 - x^2)^{n-1}}$
$\displaystyle \int \frac{dx}{x^m(a^2 - x^2)^n} = \frac{1}{a^2}\int \frac{dx}{x^m(a^2 - x^2)^{n-1}} + \frac{1}{a^2}\int \frac{dx}{x^{m-2}(a^2 - x^2)^n}$
Интеграли, съдържащи $\sqrt{x^2 + a^2}$
$\displaystyle \int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left(x + \sqrt{x^2 + a^2}\right)\allowbreak\quad\text{или}\quad\sinh^{-1}\frac{x}{a}$
$\displaystyle \int \frac{x\,dx}{\sqrt{x^2 + a^2}} = \sqrt{x^2 + a^2}$
$\displaystyle \int \frac{x^2\,dx}{\sqrt{x^2 + a^2}} = \frac{x\sqrt{x^2 + a^2}}{2} - \frac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int \frac{x^3\,dx}{\sqrt{x^2 + a^2}} = \frac{(x^2 + a^2)^{3/2}}{3} - a^2\sqrt{x^2 + a^2}$
$\displaystyle \int \frac{dx}{x\sqrt{x^2 + a^2}} = -\frac{1}{a}\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \frac{dx}{x^2\sqrt{x^2 + a^2}} = -\frac{\sqrt{x^2 + a^2}}{a^2x}$
$\displaystyle \int \frac{dx}{x^3\sqrt{x^2 + a^2}} = -\frac{\sqrt{x^2 + a^2}}{2a^2x^2} + \frac{1}{2a^3}\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \sqrt{x^2 + a^2}\,dx = \frac{x\sqrt{x^2 + a^2}}{2} + \frac{a^2}{2}\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int x\sqrt{x^2 + a^2}\,dx = \frac{(x^2 + a^2)^{3/2}}{3}$
$\displaystyle \int x^2\sqrt{x^2 + a^2}\,dx = \frac{x(x^2 + a^2)^{3/2}}{4} - \frac{a^2x\sqrt{x^2 + a^2}}{8} - \frac{a^4}{8}\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int x^3\sqrt{x^2 + a^2}\,dx = \frac{(x^2 + a^2)^{5/2}}{5} - \frac{a^2(x^2 + a^2)^{3/2}}{3}$
$\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x}\,dx = \sqrt{x^2 + a^2} - a\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2}\,dx = -\frac{\sqrt{x^2 + a^2}}{x} + \ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^3}\,dx = -\frac{\sqrt{x^2 + a^2}}{2x^2} - \frac{1}{2a}\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \frac{dx}{(x^2 + a^2)^{3/2}} = \frac{x}{a^2\sqrt{x^2 + a^2}}$
$\displaystyle \int \frac{x\,dx}{(x^2 + a^2)^{3/2}} = \frac{-1}{\sqrt{x^2 + a^2}}$
$\displaystyle \int \frac{x^2\,dx}{(x^2 + a^2)^{3/2}} = \frac{-x}{\sqrt{x^2 + a^2}} + \ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int \frac{x^3\,dx}{(x^2 + a^2)^{3/2}} = \sqrt{x^2 + a^2} + \frac{a^2}{\sqrt{x^2 + a^2}}$
$\displaystyle \int \frac{dx}{x(x^2 + a^2)^{3/2}} = \frac{1}{a^2\sqrt{x^2 + a^2}} - \frac{1}{a^3}\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \frac{dx}{x^2(x^2 + a^2)^{3/2}} = -\frac{\sqrt{x^2 + a^2}}{a^4x} - \frac{x}{a^4\sqrt{x^2 + a^2}}$
$\displaystyle \int \frac{dx}{x^3(x^2 + a^2)^{3/2}} = \frac{-1}{2a^2x^2\sqrt{x^2 + a^2}} - \frac{3}{2a^4\sqrt{x^2 + a^2}} + \frac{3}{2a^5}\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int (x^2 +\nobreak a^2)^{3/2}\,dx = \frac{x(x^2 + a^2)^{3/2}}{4} + \frac{3a^2x\sqrt{x^2 + a^2}}{8} + \frac{3}{8}a^4\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int x(x^2 +\nobreak a^2)^{3/2}\,dx = \frac{(x^2 + a^2)^{5/2}}{5}$
$\displaystyle \int x^2(x^2 +\nobreak a^2)^{3/2}\,dx = \frac{x(x^2 + a^2)^{5/2}}{6} - \frac{a^2x(x^2 + a^2)^{3/2}}{24} - \frac{a^4x\sqrt{x^2 + a^2}}{16} - \frac{a^6}{16}\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int x^3(x^2 +\nobreak a^2)^{3/2}\,dx = \frac{(x^2 + a^2)^{7/2}}{7} - \frac{a^2(x^2 + a^2)^{5/2}}{5}$
$\displaystyle \int \frac{(x^2 + a^2)^{3/2}}{x}\,dx = \frac{(x^2 + a^2)^{3/2}}{3} + a^2\sqrt{x^2 + a^2} - a^3\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
$\displaystyle \int \frac{(x^2 + a^2)^{3/2}}{x^2}\,dx = -\frac{(x^2 + a^2)^{3/2}}{x} + \frac{3x\sqrt{x^2 + a^2}}{2} + \frac{3}{2}a^2\ln\left(x + \sqrt{x^2 + a^2}\right)$
$\displaystyle \int \frac{(x^2 + a^2)^{3/2}}{x^3}\,dx = -\frac{(x^2 + a^2)^{3/2}}{2x^2} + \frac{3}{2}\sqrt{x^2 + a^2} - \frac{3}{2}a\ln\left(\frac{a + \sqrt{x^2 + a^2}}{x}\right)$
