How to integrate this $\int \frac{\sqrt{x}}{1+x^4}dx$?

Question

How to integrate this ? $$\int \frac{\sqrt{x}}{1+x^4}dx$$

Any hint or idea on how to proceed?

Edit: Here is the final answer using Wolfram Alpha

Answer 1

You can use the substitution $x=t^2$, $dx=2t\ dt$, so you get $$\int \frac{2t^2}{(1+t^8)} \ dt$$

although that one is a rational integral which involves many calculations.

To solve that rational integral, we first have to decompose $p(t)=1+t^8$ in irreducible polynomials. As it's clear $p(t)$ doesn't have real roots, its decomposition would only have polynomials of degree $2$. The easiest way to solve this is finding the $8$-th roots of $-1$, which are:

$$\xi_k=e^\frac{(2k-1)i}{8}, k\in\{0,1,...,7\}$$

If we notice that $\bar\xi_7=\xi_0$, $\bar\xi_6=\xi_1$, $\bar\xi_5=\xi_2$ and $\bar\xi_4=\xi_3$, that will quickly lead us to a factorization of $p(t)$, because

$$(t-\xi)(t-\bar\xi)=(t^2-2Re \xi +|\xi|^2)$$

As $|\xi_k|=1$, we get:

\begin{equation}\begin{split} p(t)=&(t-\xi_0)\cdots(t-\xi_7)=\\=&(t^2-2\cos \frac\pi8t+1)(t^2-2\cos \frac{3\pi}8t+1)(t^2-2\cos \frac{5\pi}8t+1)(t^2-2\cos \frac{7\pi}8t+1)=\\=&(t^2-2\cos \frac\pi8t+1)(t^2-2\cos \frac{3\pi}8t+1)(t^2+2\cos \frac{3\pi}8t+1)(t^2+2\cos \frac{\pi}8t+1 )\end{split}\end{equation}

because $\cos(\pi-\alpha)=-\cos(\alpha)$.

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(from now on, I simply explain the process. The calculations are way too messy.)

Now, we have to write $\frac{2t^2}{1+t^8}$ as a partial fractions sum, that is, finding $A,B,...H$ so that:

$$\frac{2t^2}{1+t^8}=\frac{At+B}{t^2-2\cos \frac\pi8t+1}+\frac{Ct+D}{t^2-2\cos \frac{3\pi}8t+1}+\frac{Et+F}{t^2+2\cos \frac{3\pi}8t+1}+\frac{Gt+H}{t^2+2\cos \frac{\pi}8t+1}$$

and integrate each term using the general formula (for your sanity)

$$\int\frac{mt+n}{at^2+bt+c}dt = \frac{m}{2a}\ln\left|at^2+bt+c\right|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2at+b}{\sqrt{4ac-b^2}}$$

To get the final result given by Wolfram Alpha, remember to substitute back $t=\sqrt{x}$.