Question

$\int {\tan }^{-1}xdx=$__________$+c$.

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

Answer 1

Answer: (B)

$x{\tan }^{-1}x-\frac{1}{2}log|x^2+1|$

$I=\int {\tan }^{-1}x$ dx
Here taking u$={\tan }^{-1}x$ and v$=1$
$\therefore \frac{du}{dx}=\frac{1}{1+x^2}$ and $\int vdx=\int 1dx=x$
$I=u\int vdx-\int \left(u^{\prime }\int vdx\right)dx$
(Rule of integration by parts)
$=\left({\tan }^{-1}x\right)x-\int \frac{x}{1+x^2}dx$
$=x\left({\tan }^{-1}x\right)-\frac{1}{2}\int \frac{2x}{1+x^2}dx$
$I=x{\tan }^{-1}x-\frac{1}{2}\int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx$,
Where $f\left(x\right)=1+x^2$
$=x{\tan }^{-1}x-\frac{1}{2}log|f\left(x\right)|+c$
$\therefore I=x{\tan }^{-1}x-\frac{1}{2}log|1+x^2|+c$.