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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

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4 Answers 4

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.

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

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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.

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$-\frac{1}{(1+\sqrt{x})^2}+\frac{2}{3(1+\sqrt{x})^3}+C$

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