22. $int\frac{\tan^{-1}x}{1 + x^{2}}dx$

Explanation:

Step1: Use substitution

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

Step2: Rewrite the integral

The integral $\int\frac{\tan^{-1}x}{1 + x^{2}}dx$ becomes $\int udu$.

Step3: Integrate $u$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n = 1$ in $\int udu$, we have $\frac{u^{2}}{2}+C$.

Step4: Substitute back

Substitute $u=\tan^{-1}x$ back into the result. We get $\frac{(\tan^{-1}x)^{2}}{2}+C$.

Answer:

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