Question

The integral of ${\tan }^{-1}\left(x\right)$ is -

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

Answer 1

The correct option is C $x{\tan }^{-1}\left(x\right)-\frac{1}{2}\ln |1+x^2|+c$

$\int ta{n}^{-1}\left(x\right)dx=\int \left(ta{n}^{-1}\right(x)\times 1)dx$
=$ta{n}^1\left(x\right)\int 1dx-\int \left[\frac{d}{dx}\right(ta{n}^1\left(x\right)\int 1dx]dx$
=$ta{n}^{-1}\left(x\right)x-\int \frac{1}{1+x^2}xdx$
=$xta{n}^{-1}\left(x\right)-\frac{1}{2}\int \frac{2x}{1+x^2}dx$
In the second term, if substitute, $t=1+x^2$ , $dt=2xdx$, which is the numerator. So we get
$\int ta{n}^{-1}\left(x\right)dx=xta{n}^{-1}\left(x\right)-\frac{1}{2}\int \frac{dt}{t}$
=$xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|t|$
Substituting back t = 1+$x^2$
$\int ta{n}^{-1}\left(x\right)dx=xta{n}^{-1}\left(x\right)-\frac{1}{2}ln|1+x^2|+c$