Question

Evaluate: $\int \frac{x{\tan }^{-1}x}{{\left(1+x^2\right)}^{3/2}}\text{dx}\text{=}$

• A. $\frac{\text{x + ta}{\text{n}}^{\text{- 1}}\text{x}}{\sqrt{\text{1 +}{\text{x}}^{\text{2}}}}\text{+ c}$
• B. $\frac{\text{x - ta}{\text{n}}^{\text{- 1}}\text{x}}{\sqrt{\text{1 +}{\text{x}}^{\text{2}}}}\text{+ c}$
• C. $\frac{\text{ta}{\text{n}}^{\text{- 1}}\text{x - x}}{\sqrt{\text{1 +}{\text{x}}^{\text{2}}}}\text{+ c}$
• D. None of these

Answer 1

The correct option is B $\frac{\text{x - ta}{\text{n}}^{\text{- 1}}\text{x}}{\sqrt{\text{1 +}{\text{x}}^{\text{2}}}}\text{+ c}$

$I=\int \frac{x{\tan }^{-1}x}{(1+x^2)1/2}dx$

Let ${\tan }^{-1}x=t$

$\frac{1}{1+x^2}dx=dt$

$\therefore I=\int \frac{t.\tan t}{(1+t^2)1/2}dt$

$=\int \frac{t\sin t./\cos t.}{\sec t}dt$

$=\int t.\sin t.dt$

$=-t\cos t+\sin t$

Now substituting $t={\tan }^{-1}x$

$I={\tan }^{-1}x\cos \left({\tan }^{-1}x\right)+\sin \left({\tan }^{-1}x\right)+c$

$=-{\tan }^{-1}x\times \cos \left({\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right)+\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)+c$

$=\frac{-{\tan }^{-1}x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+c$

$=\frac{x-{\tan }^{-1}x}{\sqrt{1+x^2}}+c$