Please help me to evaluate the integral: $\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$
Thanks.
Please help me to evaluate the integral: $\displaystyle {{\int_{-1}^{\infty }{\left( \frac{{{x}^{4}}}{1+{{x}^{6}}} \right)}}^{2}}dx$
Thanks.
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..
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.
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} $
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 $u = \tan\left(\theta\right)$
\begin{align} {1 \over 3}\int{\tan^{2}\left(\theta\right) \over \sec^{4}\left(\theta\right)}\, \sec^{2}\left(\theta\right)\,{\rm d}\theta &= {1 \over 3}\int\sin^{2}\left(\theta\right)\,{\rm d}\theta = {1 \over 3}\int{1 - \cos\left(2\theta\right) \over 2}\,{\rm d}\theta \\&= {1 \over 6}\,\theta - {1 \over 12}\,\sin\left(2\theta\right) = {1 \over 6}\,\theta - {1 \over 6}\,{\tan\left(\theta\right)\over \tan^{2}\left(\theta\right) + 1} \\&= {1 \over 6}\left\lbrack \arctan\left(u\right) - {u \over u^{2} + 1} \right\rbrack = {1 \over 6}\left\lbrack \arctan\left(x^{3}\right) - {x^{3} \over x^{6} + 1} \right\rbrack \end{align}