Question

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

(a) $\frac{1}{1+x^2}+C$

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

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

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

Answer 1

Let $I=\int 1 \cdot \tan ^{-1} x d x$

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

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

So, The correct option of this question will be (D).