Question

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

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

Answer 1

Answer: (A)

$I=\underset{0}{\overset{1}{\int }}\frac{\sqrt{x}}{(1+x)(1+3x)(3+x)}dx$
Let $x=t^2\Rightarrow dx=2t.dt$
$I=\underset{0}{\overset{1}{\int }}\frac{t(2t)}{\left(t^2+1\right)\left(1+3t^2\right)\left(3+t^2\right)}dt$
$I=\underset{0}{\overset{1}{\int }}\frac{\left(3t^2+1\right)-\left(t^2+1\right)}{\left(3t^2+1\right)\left(t^2+1\right)\left(3+t^2\right)}dt$
$I=\underset{0}{\overset{1}{\int }}\frac{dt}{\left(t^2+1\right)\left(3+t^2\right)}-\underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)\left(3+t^2\right)}$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{\left(3+t^2\right)-\left(t^2+1\right)}{\left(t^2+1\right)\left(3+t^2\right)}dt+\frac{1}{8}\underset{0}{\overset{1}{\int }}\frac{\left(1+3t^2\right)-3\left(3+t^2\right)}{\left(1+3t^2\right)\left(3+t^2\right)}dt$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{1+t^2}-\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+3}+\frac{1}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+3}-\frac{3}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{\left(1+3t^2\right)}$
$=\frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+1}-\frac{3}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{t^2+3}-\frac{3}{8}\underset{0}{\overset{1}{\int }}\frac{dt}{1+3t^2}$
$=\frac{1}{2}{\left({\tan }^{-1}(t)\right)}_0^1-\frac{3}{8\sqrt{3}}{\left({\tan }^{-1}\left(\frac{t}{\sqrt{3}}\right)\right)}_0^1$
$-\frac{3}{8\sqrt{3}}{\left({\tan }^{-1}(\sqrt{3}t)\right)}_0^1$
$=\frac{1}{2}\left(\frac{\pi }{4}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi }{6}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi }{3}\right)$
$=\frac{\pi }{8}-\frac{\sqrt{3}}{16}\pi$
$=\frac{\pi }{8}\left(1-\frac{\sqrt{3}}{2}\right)$