use integration by parts to evaluate the integral: $int\tan^{-1}x dx$

Explanation:

Step1: Choose $u$ and $dv$

Let $u = \tan^{- 1}x$, $dv=dx$. Then $du=\frac{1}{1 + x^{2}}dx$, $v = x$.

Step2: Apply integration - by - parts formula

The integration - by - parts formula is $\int u;dv=uv-\int v;du$. Substituting $u$, $v$, $du$ and $dv$ we get: $\int\tan^{-1}x;dx=x\tan^{-1}x-\int\frac{x}{1 + x^{2}}dx$.

Step3: Evaluate $\int\frac{x}{1 + x^{2}}dx$

Let $t = 1 + x^{2}$, then $dt = 2x;dx$, and $\int\frac{x}{1 + x^{2}}dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\ln|t|+C=\frac{1}{2}\ln(1 + x^{2})+C$.

Answer:

$x\tan^{-1}x-\frac{1}{2}\ln(1 + x^{2})+C$