Antiderivative of $\frac{\sqrt{4-x}}{x\sqrt{x}}$

Question

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 ?

Answer 1

HINT:

As $4\ge x\ge0$ write $x=4\sin^2\theta$

When $x=0,\theta=0;x=4,\theta=\pm\dfrac\pi2$

If we choose $x=4,\theta=+\dfrac\pi2,\sqrt x=2\sin\theta,\sqrt{4-x}=2\cos\theta$

If we choose $x=4,\theta=-\dfrac\pi2,\sqrt x=-2\sin\theta,\sqrt{4-x}=2\cos\theta$(why?)

Answer 2

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

$$\int\frac1x\sqrt\frac{4-x}x\;dx=\int\frac{u^2+1}4\cdot u\cdot \left(-\frac{8u}{(u^2+1)^2}du\right)=-2\int\frac{u^2}{(u^2+1)^2}du=$$

$$=-\arctan u+\frac u{u^2+1}+C\longrightarrow-\arctan\sqrt\frac{4-x}x+\frac{\sqrt\frac{4-x}x}{1+\frac{4-x}x}=$$

$$=\frac{\sqrt{x(4-x)}}4-\arctan\sqrt\frac{4-x}x+C$$