Question

If $\int \frac{\sqrt{x}}{x(x+1)}dx=kta{n}^{-1}m+C$, then $(k,m)$ is__________

• A. $(1,\sqrt{x})$
• B. $(2,\sqrt{x})$
• C. $(2,x)$
• D. $(1,x)$

Answer 1

Answer: (B)

$\int \frac{\sqrt{x}}{x(x+1)}dx=k{\tan }^{-1}m$
Put $\{\begin{array}{l}x={\tan }^2\theta \\ dx=2\tan \theta \cdot {\sec }^2\theta d\theta \end{array}$
$=\int \frac{\tan \theta }{{\tan }^2\theta \cdot {\sec }^2\theta }\cdot \left(2\tan \theta \cdot {\sec }^2\theta \right)d\theta$
$=2\int d\theta =2\theta =2{\tan }^{-1}\sqrt{x}=k{\tan }^{-1}(m)$
On comparing, we get $(k,m)=(2,\sqrt{x})$