How does my book get from this step to this step?

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[SOLVED] How does my book get from this step to this step?

My book went from int of dx/ ( sqrt(x)*(x+1) )

to

2arctan sqrt (x) -- WTH?

I thought the form for arctan was dx/a^2 + u^2 and I don't see that in the given integrand?

Can anyone help?
 
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Substitute u=\sqrt{x}, and see what you get.
 
If x=u2, then dx = 2 u du

sqrt (x) = u, and

x+1 = u2 + 1
 
astronuc, did you arrive at that by using the substitution arildno suggested?
 
Well, arildno said "u= \sqrt{x}" and Astronuc said "x= u2". Look the same to me!
 
Ha, you rascal!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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