<s>[INST] definite integral- how to solve. I know the substitution. I know the substitution but how should I continue? 
$$\int_0^1 x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$$
 [/INST] An alternative way - by setting $x=u^2$, then $u=\frac{1-v}{1+v}$ and $v=z^2$, we are left with:
$$ \int_{0}^{1}2u^3\sqrt{\frac{1-u}{1+u}}\,du = 4\int_{0}^{1}\frac{(1-v)^3}{(1+v)^5}\sqrt{v}\,dv =8\int_{0}^{1}\frac{z^2(1-z^2)^3}{(1+z^2)^5}\,dz$$
Then by applying integration by parts we get
$$ \int_{0}^{1}\frac{1-9 z^2+15 z^4-7 z^6}{(1+z^2)^4}\,dz=-7\,I_1+36\,I_2-60\,I_3+32\,I_4 $$
with $I_k=\int_{0}^{1}\frac{dz}{(1+z^2)^k}.$ These integrals can be computed through differentiation under the integral sign, since for any $a>0$ we have
$$ \int_{0}^{1}\frac{a\,dz}{a+z^2}=\frac{\text{arccot}\sqrt{a}}{\sqrt{a}} $$
hence $I_1=\frac{\pi}{4}, I_2=\frac{1}{4}+\frac{\pi}{8}, I_3=\frac{1}{4}+\frac{3\pi}{32}, I_4=\frac{11}{48}+\frac{5\pi}{64}$. Summarizing,

$$ \int_{0}^{1}x\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx = \color{red}{\frac{4}{3}-\frac{3\pi}{8}}.$$

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