Question

Integrate the function ${\tan }^{-1}x$

Answer 1

Let $I=\int 1\cdot {\tan }^{-1}xdx$
Taking ${\tan }^{-1}$ as first function and 1 as second function and integrating by parts, we obtain
$I={\tan }^{-1}x\int 1dx-\int \{\left(\frac{d}{dx}{\tan }^{-1}x\right)\int 1\cdot dx\}dx$
$={\tan }^{-1}x\cdot x-\int \frac{1}{1+x^2}\cdot xdx$
$=x{\tan }^{-1}x-\frac{1}{2}\int \frac{2x}{1+x^2}dx$
$=x{\tan }^{-1}x-\frac{1}{2}\log |1+x^2|+C$
$=x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)+C$