Question

Evaluate:$\int {\tan }^{-1}xdx$

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

Answer 1

The correct option is D

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

Evaluating the integral$\int {\tan }^{-1}xdx$

$$\begin{gathered} \Rightarrow \int \left({\tan }^{-1}x\right)1dx \\ \Rightarrow {\tan }^{-1}x\int 1dx-\int \left[\frac{d}{dx}{\tan }^{-1}x\int 1dx\right]dx\left[\text{Applying integration by parts}\right] \\ \Rightarrow x{\tan }^{-1}x-\int \frac{x}{1+x^2}dx \end{gathered}$$

Substituting $1+x^2=t$and put the value of $dx$after differentiating w.r.t. $x$

$dx=\frac{dt}{2x}$

$$\begin{gathered} \Rightarrow x{\tan }^{-1}x-\frac{1}{2}\int \frac{1}{t}dt+c \\ \Rightarrow x{\tan }^{-1}x-\frac{1}{2}\log \left|t\right|+c \\ \Rightarrow x{\tan }^{-1}x-\frac{1}{2}\log |1+x^2|+c[\text{substituting}t=(1+x^2\left)\right] \end{gathered}$$

Hence, option D is the correct answer.