Question

The value of the integral $\int\limits_{0}^{1} \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}$ 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)

$I=\int\limits_{0}^{1} \frac{\sqrt{x}}{(1+x)(1+3 x)(3+x)} d x$
Let $x=t^{2} \Rightarrow d x=2 t . d t$
$ I =\int\limits_{0}^{1} \frac{ t (2 t )}{\left( t ^{2}+1\right)\left(1+3 t ^{2}\right)\left(3+ t ^{2}\right)} dt$
$ I =\int\limits_{0}^{1} \frac{\left(3 t ^{2}+1\right)-\left( t ^{2}+1\right)}{\left(3 t ^{2}+1\right)\left( t ^{2}+1\right)\left(3+ t ^{2}\right)} dt $
$ I =\int\limits_{0}^{1} \frac{ dt }{\left( t ^{2}+1\right)\left(3+ t ^{2}\right)}-\int\limits_{0}^{1} \frac{ dt }{\left(1+3 t ^{2}\right)\left(3+ t ^{2}\right)}$
$=\frac{1}{2} \int\limits_{0}^{1} \frac{\left(3+t^{2}\right)-\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(3+t^{2}\right)} dt +\frac{1}{8} \int\limits_{0}^{1} \frac{\left(1+3 t^{2}\right)-3\left(3+t^{2}\right)}{\left(1+3 t^{2}\right)\left(3+t^{2}\right)} dt$
$=\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{1+t^{2}}-\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{t^{2}+3}+\frac{1}{8} \int\limits_{0}^{1} \frac{d t}{t^{2}+3}-\frac{3}{8} \int\limits_{0}^{1} \frac{d t}{\left(1+3 t^{2}\right)} $
$=\frac{1}{2} \int\limits_{0}^{1} \frac{d t}{t^{2}+1}-\frac{3}{8} \int\limits_{0}^{1} \frac{d t}{t^{2}+3}-\frac{3}{8} \int\limits_{0}^{1} \frac{d t}{1+3 t^{2}}$
$=\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) $
$=\frac{\pi}{8}-\frac{\sqrt{3}}{16} \pi$
$=\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{2}\right)$