Question

Consider $\int x{\tan }^{-1}dx=A(x^2+1){\tan }^{-1}x+Bx+C,$ where C is the constant of integration.

What is the value of B?

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{-1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{-1}{2}$

Given ,$\int x{\tan }^{-1}xdx=\int \frac{d}{dx}\left(\frac{x^2}{2}\right){\tan }^{-1}xdx$

$=\left(\frac{x^2}{2}\right){\tan }^{-1}x-\int \left(\frac{x^2}{2}\right)\frac{d}{dx}({\tan }^{-1}x)dx$

$=\left(\frac{x^2}{2}\right){\tan }^{-1}x-\int \left(\frac{x^2}{2}\right)\frac{1}{x^2+1}dx$

$=\left(\frac{x^2}{2}\right){\tan }^{-1}x-\frac{1}{2}\int \frac{x^2}{x^2+1}dx$

$=\left(\frac{x^2}{2}\right){\tan }^{-1}x-\frac{1}{2}\int (1-\frac{1}{x^2+1})dx$

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

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

Hence $B=\frac{-1}{2}$