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

then integrate after seperating both the terms.

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

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Let $u = 1 + \sqrt{x}$. Thus, we have $x = (u- 1)^{2}$ and $dx = 2(u-1)du$.

And so, the integral becomes

$$ = 2\int \frac{(u-1)du}{u^{4}} = 2 \int(u^{-3} - u^{-4})du $$

$$ = 2 \left( -\frac{u^{-2}}{2} + \frac{u^{-3}}{3} \right) = \frac{2}{3u^{3}} - \frac{1}{u^{2}} + C $$

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

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