Question

The value of the integral $\underset{0}{\overset{1}{\int }}\frac{\sqrt{x}dx}{(1+x)(1+3x)(3+x)}$ is

• A. $\frac{\pi }{4}(1-\frac{\sqrt{3}}{6})$
• B. $\frac{\pi }{8}(1-\frac{\sqrt{3}}{2})$
• C. $\frac{\pi }{4}(1-\frac{\sqrt{3}}{2})$
• D. $\frac{\pi }{8}(1-\frac{\sqrt{3}}{6})$

Answer 1

The correct option is B $\frac{\pi }{8}(1-\frac{\sqrt{3}}{2})$

Put $x=t^2$
$\underset{0}{\overset{1}{\int }}\frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)}dt$
$\frac{2t^2}{(1+t^2)(1+3t^2)(3+t^2)}=\frac{A}{1+t^2}+\frac{B}{1+3t^2}+\frac{C}{3+t^2}$
$\Rightarrow A=\frac{1}{2},B=-\frac{3}{8},C=-\frac{3}{8}$
$\Rightarrow \frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}-\frac{3}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{1+3t^2}-\frac{3}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{3+t^2}$
$=\frac{\pi }{8}-\frac{3}{8}(\frac{1}{\sqrt{3}}\cdot \frac{\pi }{3})-\frac{3}{8}(\frac{1}{\sqrt{3}}\cdot \frac{\pi }{6})$
$=\frac{\pi }{8}-\frac{\sqrt{3}}{16}\pi =\frac{\pi }{8}(1-\frac{\sqrt{3}}{2})$