I have question about one integral

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The Attempt at a Solution



Thanks in advance for your help...


∫[1/(1+x^2)^2] dx
 
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What did you try?

Did you try integration by parts, substitution?
 
I managed to get it by using substitution and then parts. Try making u=1/(1+x^2) and then when you get du, use u=1/(1+x^2) to solve for x to replace x with u. After that, use parts. Hope that helps I can give a more detailed answer if you still are having trouble.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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