Answer
True
Work Step by Step
True.
Let $u=\tan^{-1}(x)$ and $dv=dx$.
Then, $du=\frac{1}{x^2+1}dx$ and $v=x$.
Using the Integration by Parts,
$\int \tan^{-1}x dx=\tan^{-1}x\cdot x-\int x\cdot \frac{1}{x^2+1}dx$
$=x\tan^{-1}x-\int \frac{x}{x^2+1}dx$
$=x\tan^{-1}x-\frac{1}{2}\ln |x^2+1|+C$