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
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
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.
Consider letting $u = 1 + \sqrt{x}$ and then noting that $2 \sqrt{x} = 2(u-1)$.
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.
$-\frac{1}{(1+\sqrt{x})^2}+\frac{2}{3(1+\sqrt{x})^3}+C$