Найти интеграл: \( \int \ln {(x^2 + 1)} dx .\)

Решение

\( \int \ln {(x^2 + 1)} dx = \begin{vmatrix}
u = \ln{(x^2+1)}\\
du = \frac{1}{x^2+1}\cdot 2^x dx\\
dv = v' dx\\
x = v\\
\end{vmatrix} = \ln{(x^2 + 1)} x - \int x\cdot \frac{2dx\cdot x}{x^2+1} = \)

\(= \ln {(x^2 + 1)} x - \int\frac{2x\cdot x}{x^2 + 1} dx = \ln {(x^2+1)}x - \int \frac{2x^2}{x^2+1} dx=\)

\( = \ln{(x^2+1)}x-2\int \frac{x^2}{x^2+1} dx = \ln{(x^2+1)}x - 2\frac{x^2+1-1}{x^2+1} dx = \)

\(= \ln{(x^2+1)}x - 2\int 1 - \frac{1}{x^2+1} dx = \ln{(x^2 + 1)} x - 2(x-\arctan{x})+C=\)

\(= \ln{(x^2+1)}x - 2x + 2 \arctan{x} + C \)

Ответ: \( \ln{(x^2+1)}x - 2x + 2 \arctan{x} + C \).