# Solve${{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$

Please help me to evaluate the integral: $\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$

Thanks.

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Notice that: $$x^8 = (x^6 + 1)x^2 - x^2$$ So that your integrand takes the form: $$\int\frac{x^2}{1+x^6}dx - \int\frac{x^2}{(1+x^6)^2}dx$$ Now substitute $u = x^3, du = 3 x^2 dx$: $$\int\frac{1}{3(1+u^2)}du - \int\frac{1}{3(1+u^2)^2}du$$ The first integral is a multiple of $\arctan(u)$, and the second can be solved dividing again into two parts: $$\int\frac{1}{3(1+u^2)^2}du = \int\frac{1}{3(1+u^2)}du - \int\frac{u^2}{3(1+u^2)^2}du$$ I think you get the idea..

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Thank you very much for the help. I am able to solve it now. –  Sleepingip Dec 8 '12 at 15:22

Try substitutions of the form $u = x^k$, where $k$ should divide the exponent 6, and work on each of them. At least one of these will look much more familiar. Then work on that one, using the usual tricks.

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Here is a result by Maple for indefinite integral. I managed to get the more compact form

$$\displaystyle {{\int{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx={\frac {\arctan \left( {x}^{3} \right) {x}^{6}+\arctan \left( {x}^{3}\right) -{x}^{3}}{6\,{x}^{6}+6}}$$

The final result of the definite integral by maple is

$$-\frac{1}{12}+\frac{\pi}{8}$$

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Here is a more compact form for the answer. –  Mhenni Benghorbal Dec 8 '12 at 14:57
Thank you very much, the answer I obtain is the same with yours. –  Sleepingip Dec 8 '12 at 15:22
@Sleepingip: You are welcome. –  Mhenni Benghorbal Dec 8 '12 at 20:29
With $x = u^{1/3}$ we get
$$\int\left(x^{4} \over 1 + x^{6}\right)^{2}\,{\rm d}x = {1 \over 3}\int{u^{2} \over \left(1 + u^{2}\right)^{2}}\,{\rm d}u$$
Set $u = \tan\left(\theta\right)$