Question

If $\int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)dx=A(x){\tan }^{-1}(\sqrt{x})+B(x)+C$, where $C$ is a constant of integration, then the ordered pair $(A(x),B(x))$ can be :

• A. $(x-1,\sqrt{x})$
• B. $(x+1,\sqrt{x})$
• C. $(x+1,-\sqrt{x})$
• D. $(x-1,-\sqrt{x})$

Answer 1

Answer: (C)

Put $x={\tan }^2\theta$
$\Rightarrow dx=2\tan \theta {\sec }^2\theta d\theta$
$\int \theta .\left(2\tan \theta \cdot {\sec }^2\theta \right)d\theta$
$\downarrow \downarrow$
$III$ (By parts)
$=\theta \cdot {\tan }^2\theta -\int {\tan }^2\theta d\theta$
$=\theta \cdot {\tan }^2\theta -\int \left({\sec }^2\theta -1\right)d\theta$
$=\theta \left(1+{\tan }^2\theta \right)-\tan \theta +C$
$={\tan }^{-1}(\sqrt{x})(1+x)-\sqrt{x}+C$