Неопределени интеграли

[Гост]

а) [tex]\int[/tex] [tex]\frac{4x - arctg^{3}(5x)}{1+25x^{2}}[/tex]

б) [tex]\int[/tex] [tex]\frac{cos(x)}{1+3cos^{2}x}[/tex]

Благодаря!

[Davids]

a) $I = \int\frac{4x - arctg^3(5x)}{1 + 25x^2}dx = \int\frac{4x}{1 + 25x^2}dx - \int\frac{arctg^3(5x)}{1 + 25x^2}dx$

Първо: $I_1 = \int\frac{4x}{1 + 25x^2}dx$

Полагаме $t = 1 + 25x^2 \Rightarrow dt = 50xdx$

Тогава $I_1 = \frac{2}{25}\int\frac{1}{t}dt = \frac{2}{25}ln(t) = \frac{2}{25}ln(1 + 25x^2) + C$

Второ: $I_2 = \int\frac{arctg^3(5x)}{1 + 25x^2}dx$

Полагаме $t = arctg(5x) \Rightarrow dt = \frac{5}{1 + 25x^2}dx$

Тогава $I_2 = \frac{1}{5}\int t^3dt = \frac{1}{20}t^4 = \frac{1}{20}arctg^4(5x) + C$

Окончателно $I = I_1 - I_2 = \frac{2}{25}ln(1 + 25x^2) - \frac{1}{20}arctg^4(5x) + C$

[Davids]

б) $I = \int\frac{cosx}{1 + 3cos^2x}dx = \int\frac{cosx}{4 - 3sin^2x}dx$

Полагаме $t = sinx \Rightarrow dt = cosxdx$

Тогава $I = \frac{1}{4 - 3t^2}dt = \frac{1}{3}\int\frac{1}{\frac{4}{3} - t^2}dt = \frac{1}{3}\int\frac{A}{\frac{2}{\sqrt{3}} - t} + \frac{B}{\frac{2}{\sqrt{3}} + t}dt$

Сега искаме $A(\frac{2}{\sqrt{3}} + t) + B(\frac{2}{\sqrt{3}} - t) = 1 \Leftrightarrow \begin{array}{|l} \frac{2}{\sqrt{3}}(A + B) = 1 \\ A - B = 0 \end{array} \Rightarrow A = B = \frac{\sqrt{3}}{4}$

Значи $I = \frac{1}{4\sqrt{3}}\int\frac{1}{\frac{2}{\sqrt{3}} - t} + \frac{1}{\frac{2}{\sqrt{3}} + t}dt = \frac{1}{4\sqrt{3}}\Bigg(-ln\Big|\frac{2}{\sqrt{3}} - t\Big| + ln\Big|\frac{2}{\sqrt{3}} + t\Big|\Bigg) = \frac{1}{4\sqrt{3}}ln\Big|\frac{2\sqrt{3} + 3t}{2\sqrt{3} - 3t}\Big| = \frac{1}{4\sqrt{3}}ln\Bigg(\frac{2\sqrt{3} + 3sinx}{2\sqrt{3} - 3sinx}\Bigg) + C$