TITLE: Complex Definite Integral: $\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$
QUESTION [8 upvotes]: Compute the following definite integral:
$$\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$$
This is what I did:
$$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4} \, dx$$
$$u = \sqrt{x}$$
$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}} \, dx$$
$$du = \frac{1}{2\sqrt{x}} \, dx$$
And after this I just got stuck. How exactly am I supposed to write $du$ in terms of the initial integral? I can't double it and nor can I leave it as is because of the $+1$. Am I supposed to make $u = 1 + \sqrt{x}$ instead or is there a way to do it with the current $du$?
Any help?

REPLY [4 votes]: You have $du = \dfrac {dx}{2\sqrt x},$ and from that you get $du = \dfrac{dx}{2u}$ so that $2u\,du = dx.$
But a quicker way is to differentiate both sides of $u^2=x$ to get $2u\,du = dx.$
Notice that as $x$ goes from $0$ to $1,$ so does $u,$ so the bounds of integration do not change.
$$
\int_0^1 \frac{dx}{(1+\sqrt x\,)^4} = \int_0^1 \frac{2u\,du} {(1+u)^4} = \int_0^1 \left( \frac A {1+u} + \frac B {(1+u)^2} + \frac C {(1+u)^3} + \frac D {(1+u)^4} \right) \, du.
$$
You need to do some algebra to find $A,B,C,D.$