Does this question require double u-substitution?

Question

$$\int_0^1\frac1{(1+\sqrt x)^4}\,dx$$ The first substitution I thought to do was $u=\sqrt x$ and $du=\frac 12x^{-1/2}\,dx$. So what I have my integral change into is $$\int_0^1\frac1{(1+\sqrt u)^4}\,du$$ I am not sure how to use the $du$ and where to go from there.

Answer 1

The substitution was performed wrongly. I would suggest using $u=1+\sqrt x$ with $\frac{du}{dx}=\frac1{2\sqrt x}$: $$=\int_1^2\frac{2(u-1)}{u^4}\,du$$ $$=2\int_1^2\left(\frac1{u^3}-\frac1{u^4}\right)\,du$$ $$=2\left[-\frac1{2u^2}+\frac1{3u^3}\right]_1^2$$ $$=2\left(-\frac18+\frac1{24}+\frac12-\frac13\right)=\frac16$$ So only a single substitution is required, in fact.

Answer 2

HINT: Let $x=u^2$. Then, $dx=2udu$.

$$\int_0^1\frac{1}{(1+\sqrt x)^4}dx=\int_0^1\frac{2u}{(1+u)^4}du$$

Answer 3

Or: $$1+\sqrt{x}=u \Rightarrow x=(u-1)^2 \Rightarrow dx=2(u-1)du.$$ $$\int_1^2 \frac{2(u-1)du}{u^4}=\int_1^2 2u^{-3}du -\int_1^2 2u^{-4}du=-u^{-2}|_1^2+\frac23u^{-3}|_1^2=...$$