Question

$\int \frac{1}{(16+x^2{)}^{\frac{3}{2}}}dx=$

• A. $\frac{1}{16}\sin \left(\frac{1}{4}{\tan }^{-1}x\right)+C$
• B. $-\frac{1}{16}\sin \left(\frac{1}{4}{\tan }^{-1}x\right)+C$
• C. $\frac{1}{16}\sin \left(\frac{1}{5}{\tan }^{-1}x\right)+C$
• D. $\frac{1}{16}\sin \left(\frac{1}{8}{\tan }^{-1}x\right)+C$

Answer 1

The correct option is A $\frac{1}{16}\sin \left(\frac{1}{4}{\tan }^{-1}x\right)+C$

$\int \frac{1}{(16+x^2{)}^{\frac{3}{2}}}dx$

$x=4\tan \theta$ , $\theta =\frac{1}{4}{\tan }^{-1}x$

$dx=4{\sec }^2\theta d\theta$

$\int \frac{4{\sec }^2\theta d\theta }{(16{\tan }^2\theta +16{)}^{\frac{3}{2}}}$

$\frac{4}{64}\int \frac{{\sec }^2\theta d\theta }{(1+{\tan }^2\theta {)}^{\frac{3}{2}}}$

$=\frac{1}{16}\int \frac{{\sec }^2\theta d\theta }{({\sec }^2\theta {)}^{\frac{3}{2}}}$

$=\frac{1}{16}\int \frac{d\theta }{\sec \theta }$

$=\frac{1}{16}\int \cos \theta d\theta$

$=\frac{1}{16}\sin \theta +C$

$=\frac{1}{16}\sin \left(\frac{1}{4}{\tan }^{-1}x\right)+C$