Complex Definite Integral: $\int _0^1\frac{dx}{\left(1+\sqrt{x}\right)^4}$

Question

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?

Answer 1

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

$$\begin{align} \int\frac1{(1+\sqrt x)^4}\,dx&=2\int \frac{u}{(1+u)^4}\,du\\\\ &=2\int \frac{t-1}{t^4}\,dt\\\\ &=\frac{2}{3t^3}-\frac{1}{t^2}+C\\\\ &=\frac{2}{3(1+u)^3}-\frac{1}{(1+u)^2}+C\\\\ &=\frac{2}{3(1+\sqrt x)^3}-\frac{1}{(1+\sqrt x)^2}+C \end{align}$$

Answer 2

The answer is $$\int_0^1 \frac{dx}{(1+\sqrt x\,)^4} = 1/6$$

Let $$x=(t-1)^2$$ $$dx=2(t-1)dt$$ $$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx =$$

$$\int _1^{2}\frac {2(t-1)dt}{t^4}=$$

$$2\int _1^{2} {(t^{-3}-t^{-4})dt}= \frac {1}{6}$$

Answer 3

Substitute $u=1+\sqrt x$ and $ du=\frac {dx} {2\sqrt x}=\frac {dx} {2(u-1)}$

$$\int _0^1\frac{1}{\left(1+\sqrt{x}\right)^4}dx=2\int _1^2\frac{\sqrt xdu} {u^4}=2\int _1^2\frac{u-1} {u^4}du=2\int_1^2\frac{du} {u^3}-2\int_1^2\frac{du} {u^4}$$

Answer 4

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 $A,B,C,D.$

Answer 5

Let $\sqrt{x}=\tan ^2 \theta$, then $$ \begin{aligned} \int_0^1 \frac{1}{(1+\sqrt{x})^4} d x & =4 \int_0^ {\frac{\pi}{4}} \sin ^3 \theta \cos ^3 \theta d \theta \\ & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \sin ^3 2\theta d \theta \\ & =-\frac{1}{4} \int_0^{\frac{\pi}{4}}\left(1-\cos ^2 2 \theta\right) d(\cos 2 \theta) \\ & =-\frac{1}{4}\left[\cos 2 \theta-\frac{\cos ^3 2 \theta}{3}\right]_0^{\frac{\pi}{4}} \\ & =\frac{1}{6} \end{aligned} $$