<s>[INST] Antiderivative of $\frac{\sqrt{4-x}}{x\sqrt{x}}$ I need help to find the antiderivative of the function $\displaystyle x \, \mapsto \, \frac{\sqrt{4-x}}{x\sqrt{x}}$ on $]0,4[$. I have tried the change of variables $u = \sqrt{4-x}$ but it didn't help. Which change of variables (or trick) am I not thinking of ?
 [/INST] Let us try this:
$$\frac{4-x}x=u^2\implies x=\frac4{u^2+1}\implies dx=-\frac{8u}{(u^2+1)^2}\,du$$
and then
$$\int\frac1x\sqrt\frac{4-x}x\;dx=\int\frac{u^2+1}4\cdot u\cdot \left(-\frac{8u}{(u^2+1)^2}du\right)=-2\int\frac{u^2}{(u^2+1)^2}du=$$
$$=-\arctan u+\frac u{u^2+1}+C\longrightarrow-\arctan\sqrt\frac{4-x}x+\frac{\sqrt\frac{4-x}x}{1+\frac{4-x}x}=$$
$$=\frac{\sqrt{x(4-x)}}4-\arctan\sqrt\frac{4-x}x+C$$
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