Jee Calculus problem

$\text{Motivation}:$ After thinking for a while and looking at the problem this general form comes to mind: $\int\dfrac{dx}{\sqrt{a^2-x^2}}$ because of the presence of the square root. $$ Now,we want to convert $(x+1)(3-x)$ into the form $a^2-x^2$ ,we can do this by writing one of the two factors in the from $(a-z)$ and the other in the form $(a+z)$ where $z$ is a linear function in $x$ and $a$ a constant.We see that the sum of the two factors is $2a=4\longrightarrow a=2$ hence we get that $z=x-1$ because then the two factors would become $z+2$ and $2-z$ and $dx=dz$ ,so the integral becomes, $\int_{-2}^{2}\dfrac{dz}{\sqrt{2^2-z^2}}=\sin^{-1}(\dfrac{z}{2})|_{-2}^{2}\\ =2\times\sin^{-1}1=2\times\dfrac{\pi}{2}=\pi$ .And done!