Question

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

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

Answer 1

Answer: (B)

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