The integral $\int \tan ^ { - 1 } x d x$ is

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

Question

The integral $\int \tan ^ { - 1 } x d x$ is

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

Quizplus · Accepted Answer

The integral of $\tan^{-1}x$ can be solved using integration by parts, where $u = \tan^{-1}x$ and $dv = dx$. The result is $\frac{2x\tan^{-1}x - \ln|x^2+1|}{2} + C$, which matches option A.

The Integral ∫tan⁡−1xdx\int \tan ^ { - 1 } x d x∫tan−1xdx

The Integral $\int \tan ^ { - 1 } x d x$