Answer
$\dfrac{\pi}{2}$
Work Step by Step
We are given that $f(x)=\int_0^1 \dfrac{dx}{\sqrt x(x+1)}.....(1)$
Let us consider that $a=\sqrt x$ and $x=a^2 \implies dx=2a da$
So, the equation (1) becomes:
$f(x)= 2 \int_0^1 \dfrac{da}{a^2+1}=2 tan^{-1} (a) +C|_0^1\\= 2 \tan^{-1} \sqrt x+ C|_0^1\\=2 \lim\limits_{b \to 0^{+}}\tan^{-1} \sqrt x|_b^1 \\=\dfrac{\pi}{2}$
