Show that $\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}$ Prove that,
$$
\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}
$$
What kind of subsititution should be used to solve this integral
Another integral that give the same answer but with a different limit and same denominator?
$$
\int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}
$$
 A: If we make the substitution $t=\frac1x$ and later $x^{10}=u$, then we find that
$$\begin{align}\int_0^1\frac{1+x^8}{1+x^{10}}dx&=\int_{\infty}^1\frac{1+t^{-8}}{1+t^{-10}}\frac{(-dt)}{t^2}=\int_1^{\infty}\frac{t^8+1}{t^{10}+1}dt\\
&=\frac12\int_0^{\infty}\frac{1+x^8}{1+x^{10}}dx=\frac12\int_0^{\infty}\frac{1+u^{\frac45}}{1+u}\frac{du}{10u^{\frac9{10}}}=\\
&\frac1{20}\int_0^{\infty}\frac{u^{\frac1{10}-1}+u^{\frac9{10}-1}}{1+u}du\end{align}$$
Now,
$$\int_0^{\infty}\frac{u^{p-1}}{1+u}du=\frac{\pi}{\sin p\pi}$$
Is a famous integral that converges for $p\in(0,1)$ and may be determined using the 'keyhole' contour. Of course $\sin\left(\frac{\pi}{10}\right)=\sin\left(\frac{9\pi}{10}\right)=\cos\left(\frac{2\pi}5\right)=\frac1{2\phi}$, where $\phi=\frac{\sqrt5+1}2$. So
$$\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{2\pi}{20\sin\left(\frac{\pi}5\right)}=\frac{\pi}{\left(\frac5{\phi}\right)}$$
So we just have to compute
$$\begin{align}\phi^5-8&=\left(\frac{\sqrt5+1}2\right)^5-8=\frac{11+5\sqrt5}2-8\\
&=\frac{5\sqrt5-5}2=5\left(\frac{\sqrt5-1}2\right)\\
&=5\left(\frac{\sqrt5+1}2\right)^{-1}=\frac5{\phi}\end{align}$$
To conclude
$$\int_0^1\frac{1+x^8}{1+x^{10}}dx=\frac{\pi}{\phi^5-8}$$
Now, you can do this with partial fractions, but is seems much more long-winded.
