Question

Evaluate $\int \frac{{\tan }^{-1}x}{1+x^2}dx$ $=$

• A. $\frac{1}{2}({\tan }^{-1}x{)}^2+C$
• B. $\frac{1}{2}\left({\tan }^{-1}x\right)+C$
• C. $\frac{1}{2}(1+x^2{)}^2+C$
• D. None

Answer 1

The correct option is A $\frac{1}{2}({\tan }^{-1}x{)}^2+C$

$I=\int \left(\frac{ta{n}^{-1}x}{1+x^2}\right)dxletta{n}^{-1}x=tthen\left(\frac{1}{1+x^2}\right)dx=dtI=\int tdt=\left(\frac{t^2}{2}\right)+C=\left(\frac{(ta{n}^{-1}x{)}^2}{2}\right)+C$