Integrate $\frac{1}{(1+\sqrt{x})^4}$ via Substitution

Question

I'm trying to solve the integral:

$$\int \frac{1}{(1+\sqrt{x})^4} \mathrm{d}x$$

I know I most likely need to use some trig substitution but really have no idea where to go with this, please help?

Thanks

Answer 1

There are several ways to turn this into a nicer problem. We will deliberately make an imperfect but natural substitution.

Let $x=u^2$. Then $dx=2u\,du$. We end up having to integrate $\dfrac{2u}{(1+u)^4}$.

Make the substitution $w=1+u$. Problem collapses.

Answer 2

Consider letting $$u = 1 + \sqrt{x}$$ and then noting that $2 \sqrt{x} = 2(u-1)$.

Answer 3

Let $1 + \sqrt x = t$ then, $\frac1{\sqrt x} dx = 2\,dt$ and now $\mathcal{I} = \int \frac{2 (t-1)}{t^4}\,dt$

then integrate after seperating both the terms.