The value of the integral

$\mathrm{I}=\int_{0}^{1} \frac{\sqrt{\mathrm{x}}}{(1+\mathrm{x})(1+3 \mathrm{x})(3+\mathrm{x})} \mathrm{d} \mathrm{x}$

Let $x=t^{2} \Rightarrow d x=2 t \cdot d t$

$I=\int_{0}^{1} \frac{t(2 t)}{\left(t^{2}+1\right)\left(1+3 t^{2}\right)\left(3+t^{2}\right)} d t$

$I=\int_{0}^{1} \frac{\left(3 t^{2}+1\right)-\left(t^{2}+1\right)}{\left(3 t^{2}+1\right)\left(t^{2}+1\right)\left(3+t^{2}\right)} d t$

$I=\int_{0}^{1} \frac{d t}{\left(t^{2}+1\right)\left(3+t^{2}\right)}-\int_{0}^{1} \frac{d t}{\left(1+3 t^{2}\right)\left(3+t^{2}\right)}$

$=\frac{1}{2} \int_{0}^{1} \frac{\left(3+t^{2}\right)-\left(t^{2}+1\right)}{\left(t^{2}+1\right)\left(3+t^{2}\right)} d t+\frac{1}{8} \int_{0}^{1} \frac{\left(1+3 t^{2}\right)-3\left(3+t^{2}\right)}{\left(1+3 t^{2}\right)\left(3+t^{2}\right)} d t$

$=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{1+\mathrm{t}^{2}}-\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}+\frac{1}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\left(1+3 \mathrm{t}^{2}\right)}$

$=\frac{1}{2} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{\mathrm{t}^{2}+3}-\frac{3}{8} \int_{0}^{1} \frac{\mathrm{dt}}{1+3 \mathrm{t}^{2}}$

$=\frac{1}{2}\left(\tan ^{-1}(\mathrm{t})\right)_{0}^{1}-\frac{3}{8 \sqrt{3}}\left(\tan ^{-1}\left(\frac{\mathrm{t}}{\sqrt{3}}\right)\right)_{0}^{1}$

$-\frac{3}{8 \sqrt{3}}\left(\tan ^{-1}(\sqrt{3} t)\right)_{0}^{1}$

$=\frac{1}{2}\left(\frac{\pi}{4}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi}{6}\right)-\frac{\sqrt{3}}{8}\left(\frac{\pi}{3}\right)$

$=\frac{\pi}{8}-\frac{\sqrt{3}}{16} \pi$

$=\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{2}\right)$