Question

Solve $\int {\tan }^{-1}xdx$

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

Answer 1

The correct option is B $x$ ${\tan }^{-1}x-\frac{1}{2}\log |1+x^2|+c$

$\int {\tan }^{-1}x.1dx$

Let us assume here ${\tan }^{-1}x$ is the first function and constant 1 is the second function. Then the integral of the second function is x.

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, we get

$={\tan }^{-1}x\int 1dx$ $-\int (d\frac{ta{n}^{-1}x}{dx}\int 1dx)dx$

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

$=x{\tan }^{-1}x+c-\frac{1}{2}\int \frac{2xdx}{1+x^2}$

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

$=\int {\tan }^{-1}xdx=x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)+c$