Trouble evaluating an integral

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Indefinite Integral(arctan(x/2) / (x^2 + 4) )dx

(sorry, I can't get my word equation editor to paste it here)

I have stared at this problem for a while, but I can't figure it out. I made the triangle that has tan = x/2, and that triangle has a hypotenuse of sqrt(4 + x^2). But that doesn't seem to get me anywhere. Also, I thought using a u-substitution by u = arctan(x/2) would work, because it would get rid of the arctan, but that's not leading me anywhere either.
Could someone point me in the right direction?
 
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u=arctan(x/2) is correct approach. What does that make du?
 
That makes du = 1/ (1 + (x^2/4)) dx
But how does that cancel out the 4 + x^2 in the denominator?
I get:
Integral(u *(1 + (x^2/4)) / 4 + x^2)du
 
Nano said:
That makes du = 1/ (1 + (x^2/4)) dx
But how does that cancel out the 4 + x^2 in the denominator?
I get:
Integral(u *(1 + (x^2/4)) / 4 + x^2)du

(4+x^2)=4*(1+x^2/4). But you are also missing a factor of (1/2) in du. Don't forget the chain rule.
 
Oh! That's kind of tricky. Thank you for your help, I appreciate it!
 
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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