The value of the integral ∫ 0 1 x d x ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) is:

Let $\sqrt{x}=t$

$\Rightarrow x=t^2, dx=2tdt$

Let $I={\int }_0^1\frac{\sqrt{x}dx}{(1+x)(1+3x)(3+x)}$

$I={\int }_0^1 \frac{2t^{2}dt}{\left(t^2+1\right)\left(3t^2+1\right)\left(t^2+3\right)}$

$={\int }_0^1 \frac{\left(\left(3t^2+1\right)-\left(t^2+1\right)\right)dt}{\left(t^2+1\right)\left(3t^2+1\right)\left(t^2+3\right)}$

$={\int }_0^1 \left(\frac{1}{\left(t^2+3\right)\left(t^2+1\right)}-\frac{1}{\left(t^2+3\right)\left(3t^2+1\right)}\right)dt$

$={\int }_0^1 \left(\frac{dt}{2\left(t^2+1\right)}-\frac{1}{8}\frac{3dt}{\left(3t^2+1\right)}-\frac{3}{8}\frac{dt}{\left(t^2+3\right)}\right)$

$={\left(\frac{1}{2}{\tan }^{-1}t-\frac{3\sqrt{3}}{8\times 3}{\tan }^{-1}\sqrt{3}t-\frac{3}{8\sqrt{3}}{\tan }^{-1}\frac{t}{\sqrt{3}}\right)}_0^1$

$=\frac{\pi }{8}-\frac{\sqrt{3}}{8}\times \frac{\pi }{3}-\frac{\sqrt{3}}{8}\times \frac{\pi }{6}$

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

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