Question

$\int \frac{x^2{\tan }^{-1}x}{1+x^2}dx$

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

Answer 1

Answer: (D)

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

Let $I=\int \frac{x^2{\tan }^{-1}x}{1+x^2}dx.$
Put $x=\tan \theta dx={\sec }^2\theta d\theta$
$\therefore I=\int \frac{\left({\tan }^2\theta \right)\theta }{{\sec }^2\theta }{\sec }^2\theta d\theta =\int \theta {\tan }^2\theta d\theta$
$=\theta \tan \theta -\log \sec \theta -\frac{{\theta }^2}{2}$
$=x{\tan }^{-1}x-\log \sqrt{1+{\tan }^2\theta }-\frac{1}{2}{\left({\tan }^{-1}x\right)}^2.$
$=x{\tan }^{-1}x-\frac{1}{2}\log (1+x^2)-\frac{1}{2}{\left({\tan }^{-1}x\right)}^2.$
Hence, option 'C' is correct.