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, 1/sqrt(x) dx = 2dt and now I = Integral {2 (t-1)/t^4}dt
then integrate after seperating both the terms.
$-\frac{1}{(1+\sqrt{x})^2}+\frac{2}{3(1+\sqrt{x})^3}+C$