Complex Definite Integral: $\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$ 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?
 A: The substitution in the OP is equivalent to $x=u^2$.  Then, $dx=2u\,du$ and we have
$$\int\frac1{(1+\sqrt x)^4}\,dx=2\int \frac{u}{(1+u)^4}\,du$$
Next, enforce the substitution $1+u=t$ to find
$$\begin{align}
\int\frac1{(1+\sqrt x)^4}\,dx&=2\int \frac{u}{(1+u)^4}\,du\\\\
&=2\int \frac{t-1}{t^4}\,dt\\\\
&=\frac{2}{3t^3}-\frac{1}{t^2}+C\\\\
&=\frac{2}{3(1+u)^3}-\frac{1}{(1+u)^2}+C\\\\
&=\frac{2}{3(1+\sqrt x)^3}-\frac{1}{(1+\sqrt x)^2}+C
\end{align}$$
A: The answer is $$\int_0^1 \frac{dx}{(1+\sqrt x\,)^4} = 1/6$$
Let $$x=(t-1)^2$$
$$dx=2(t-1)dt$$
 $$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx =$$
$$\int _1^{2}\frac {2(t-1)dt}{t^4}=$$
$$2\int _1^{2} {(t^{-3}-t^{-4})dt}= \frac {1}{6}$$
A: Substitute $u=1+\sqrt x$ and $ du=\frac {dx} {2\sqrt x}=\frac {dx} {2(u-1)}$ 
$$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx=2\int _1^2\frac{\sqrt xdu} {u^4}=2\int _1^2\frac{u-1} {u^4}du=2\int_1^2\frac{du} {u^3}-2\int_1^2\frac{du} {u^4}$$
A: 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 $A,B,C,D.$
