How to integrate int -1/(x^2 +1)^2 dx

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Homework Statement



Find the following intergral...

Homework Equations



\int{\frac{-1}{(x^2+1)^2}dx}

The Attempt at a Solution



I can't get anywhere with this... any hints?
 
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sorry rushed the problem and did it wrong
 
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hint:

\frac{d}{{dx}}{\arctan{x}} = \frac{1}{x^2 +1}
 
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Try adding and subtracting x^2 in the numerator.
 
Make the substitution x=\tan t. You'll eventually get \int \cos^2 t \ dt which is easy to do.

Daniel.
 

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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