The value of the integral $\int_{0}^{1} \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}$ is:
Correct Option: 1
$\mathrm{I}=\int_{0}^{1} \frac{\sqrt{\mathrm{x}}}{(1+\mathrm{x})(1+3 \mathrm{x})(3+\mathrm{x})} \mathrm{d} \mathrm{x}$
Let $x=t^{2} \Rightarrow d x=2 t \cdot d t$
$I=\int_{0}^{1} \frac{t(2 t)}{\left(t^{2}+1\right)\left(1+3 t^{2}\right)\left(3+t^{2}\right)} d t$
$I=\int_{0}^{1} \frac{\left(3 t^{2}+1\right)-\left(t^{2}+1\right)}{\left(3 t^{2}+1\right)\left(t^{2}+1\right)\left(3+t^{2}\right)} d t$
$I=\int_{0}^{1} \frac{d t}{\left(t^{2}+1\right)\left(3+t^{2}\right)}-\int_{0}^{1} \frac{d t}{\left(1+3 t^{2}\right)\left(3+t^{2}\right)}$
$=\frac{1}{2} \int_{0}^{1} \frac{\left(3+t^{2}\right)-\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(3+t^{2}\right)} d t+\frac{1}{8} \int_{0}^{1} \frac{\left(1+3 t^{2}\right)-3\left(3+t^{2}\right)}{\left(1+3 t^{2}\right)\left(3+t^{2}\right)} d t$
$=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}-\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}+\frac{1}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\left(1+3 \mathrm{t}^{2}\right)}$
$=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{1+3 \mathrm{t}^{2}}$
$=\frac{1}{2}\left(\tan ^{-1}(\mathrm{t})\right)_{0}^{1}-\frac{3}{8 \sqrt{3}}\left(\tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{3}}\right)\right)_{0}^{1}$
$-\frac{3}{8 \sqrt{3}}\left(\tan ^{-1}(\sqrt{3} t)\right)_{0}^{1}$
$=\frac{1}{2}\left(\frac{\pi}{4}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi}{3}\right)$
$=\frac{\pi}{8}-\frac{\sqrt{3}}{16} \pi$
$=\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{2}\right)$
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