Question

The integral $\displaystyle \int tan^{- 1} x \, dx$ is equal to (where, $C$ is the constant of integration)

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

Answer 1

Answer: (D)

Let, $I=\displaystyle \int 1.tan^{- 1}xdx$
$=\left(\tan ^{-1} x\right) \cdot x-\int \frac{1}{1+x^{2}} \cdot x d x$
$I=xtan^{- 1}x-\frac{1}{2}\displaystyle \int \frac{2 x}{1 + x^{2}} d x$
$I=xtan^{- 1}x-\frac{1}{2}log\left|\right. 1 + x^{2} \left|\right.+C$