I need to evaluate the following integral : $$I=\int_0^{\infty} \frac{x^6+1}{x^{12}+1}\mathrm dx$$
Simplifying denominator, we have :
$$I=\int_0^{\infty} \frac{x^6+1}{x^4+1}.\frac{1}{x^8-x^4+1}\mathrm dx$$
The first fraction is theoretically integrable by long division. But the second fraction does not seem promising : so integration by parts cannot be used. With such high exponents, I cannot come up with any algebraic or trigonometric substitutions.
How can I approach to solve this ? Is there any particular trick here which I'm missing ?
According to answer key, $$I=\frac{\pi}{√6}$$
Best Answer
Simplify the integral first and then partially fractionalize \begin{align} &\int_0^{\infty} \frac{x^6+1}{x^{12}+1}\overset{x\to 1/x}{dx}=\int_0^{\infty} \frac{x^{10}+x^4}{x^{12}+1}{dx}\\ =&\ \frac12 \int_0^{\infty} \frac{x^{10}+x^6+x^4+1}{x^{12}+1}{dx}=\frac12 \int_0^{\infty} \frac{x^6+1}{x^8-x^4 +1}dx \\ =&\ \frac{\sqrt3+1}{4\sqrt3} \int_0^{\infty} \frac{x^2+1}{x^4+\sqrt3x^2 +1}dx+ \frac{\sqrt3-1}{4\sqrt3} \int_0^{\infty} \frac{x^2+1}{x^4-\sqrt3x^2 +1}dx \\ =&\ \frac{\sqrt3+1}{4\sqrt3} \int_0^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2+\sqrt3} + \frac{\sqrt3-1}{4\sqrt3} \int_0^{\infty} \frac{d(x-\frac1x)}{(x-\frac1x)^2+2-\sqrt3}\\ =&\ \frac{\pi}{4\sqrt{3}}\left(\frac{\sqrt3+1}{\sqrt{2+\sqrt3}}+ \frac{\sqrt3-1}{\sqrt{2-\sqrt3}} \right)=\frac\pi{\sqrt6} \end{align}