Question

Evaluate: $\int x\frac{d}{dx}\left({\tan }^{-1}x\right)dx$

• A. $\frac{1}{1+x^2}+c$
• B. $x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)+c$
• C. $\frac{\ln (1+x^2)}{2}+c$
• D. $\frac{1}{\sqrt{1+x^2}}+c$

Answer 1

Answer: (C)

$\frac{\ln (1+x^2)}{2}+c$

Lets write the integral in this way,

$I=\int x\frac{d}{dx}\left({\tan }^{-1}x\right)dxI=\int \frac{x}{1+x^2}dxsubstitute{x}^2=t\therefore 2xdx=dtI=\int \frac{x}{1+t}\frac{dt}{2x}=\frac{1}{2}\int \frac{dt}{1+t}I=\frac{1}{2}\ln (1+t)+CI=\frac{\ln (1+x^2)}{2}+C$