A general method for integration of rational function.

Question

$\int\frac {x^3}{1+x^5}$

ATTEMPT:

I did the following substitution:

Let $x=\frac{1}{t}.$

$dx=\frac{-1}{t^2}dt.$

substituting back:

$I=\int\frac{-1}{1+t^5}dt$ which doesn't seems a simpler integration.

Next i substituted $x=p^\frac{2}{5}.$

$dx=\frac{2}{5}\frac{p^{3/5}}{1+p^2}.$

Now let $p=tan\theta.$

$dp=sec^2\theta$$d\theta.$

substituting back:

$I=\frac{2}{5}\int tan^{3/5}\theta$$d\theta$ which i couldn't integrate further even by trying By parts method.

Is there a general approach for problems of the form (not the algebraic twins):

$\int \frac{x^m}{1+x^n}dx$ where $n-m \ne1 $

Answer 1

What you already made is $$I=\int\frac {x^3}{1+x^5}dx=-\int\frac {dt}{1+t^5}$$ Now $$1+t^5=\prod_{i=1}^5 (t-\alpha_i)$$ where $\alpha_i=-(-1)^{\frac i5}$ which means that, using partial fraction decomposition, $$\frac {1}{1+t^5}=\sum_{i=1}^5 \frac{\beta_i}{t-\alpha_i}$$ $$\int\frac {dt}{1+t^5}=\sum_{i=1}^5 \beta_i \log({t-\alpha_i})$$ which is simple, except that the $\alpha_i$'s and $\beta_i$'s are complex numbers.

But the real part of $$(a+i b)\log(c+id)=\frac a2\, \log(c^2+d^2)-b\, \tan^{-1} \big(\frac d c\big)$$ and this has to be applied to each term (noticing that four of the roots $\alpha_i$ are conjugated by pairs). So, isolating the case of the real root, we need to combine by pairs the logarithms and the inverse tangents.

This is how, in "Table of Integrals, Series, and Products" by I.S. Gradshteyn and I.M. Ryzhik, they arrive to equation $2.142$. $$\frac{1}{20} \left(4 \log (t+1)+\left(\sqrt{5}-1\right) \log \left(t^2+\frac{1}{2} \left(\sqrt{5}-1\right) t+1\right)-\left(1+\sqrt{5}\right) \log \left(t^2-\frac{1}{2} \left(1+\sqrt{5}\right) t+1\right)-2 \sqrt{10-2 \sqrt{5}} \tan ^{-1}\left(\frac{-4 t+\sqrt{5}+1}{\sqrt{10-2 \sqrt{5}}}\right)+2 \sqrt{ \left(10+2\sqrt{5}\right)} \tan ^{-1}\left(\frac{4 t+\sqrt{5}-1}{\sqrt{ \left(10+2\sqrt{5}\right)}}\right)\right)$$

Answer 2

There is a general closed form for such integrals but it's not elementary or pretty at all.

$$\int \frac{x^m}{1+x^n} \, \mathrm{d}x = \frac{x^{m+1} \, _2F_1\left(1, \frac{m+1}{n};\frac{m+n+1}{n}; -x^n\right)}{m+1} + \mathrm{C}.$$

Where $_2F_1\left(a, b;c; x\right)$ is the hypergeometric function.

With regards to your specific problem, try partial fractions (although I suspect you may have mistyped, given how ugly this is)

$$\frac{x^3}{1+x^5} = \frac{\sqrt{5}x - 5x + \sqrt{5} + 5}{5\sqrt{5}(2x^2 + \sqrt{5}x - x + 2)} + \frac{\sqrt{5}x + 5x + \sqrt{5} - 5}{5\sqrt{5}(2x^2 - \sqrt{5}x - x + 2)} - \frac{1}{5(x+1)}.$$

Answer 3

For the integral $$\int \frac{x^m}{x^n + 1} \,dx ,$$ the substitution $u = x^k$, $k \in \{2, 3, \ldots\}$, will only really be useful for simplifying this integral (more precisely, will not introduce any fractional powers) if $k \mid n$ and $m \equiv -1 \pmod k$.

If $n$ is prime, as in the example $m = 3, n = 5$, the only option is $k = n$, which means a substitution requires $m \equiv -1 \pmod n$, which is not the case for that example.

Once we've reached the point where no more substitutions are available, we can use that the denominator factors over $\Bbb R$ as $$x^n + 1 = (x + 1) \prod_{k = 1}^{(n - 1) / 2} \left(x^2 - 2 \cos \frac{(2 k + 1)\pi }{n} \cdot x+ 1\right)$$ if $n$ is odd and as $$x^n + 1 = \prod_{k = 1}^{n / 2} \left(x^2 - 2 \cos \frac{(2 k + 1) \pi}{n} \cdot x+ 1\right)$$ if $n$ is even.

In our running example, $n = 5$, so $$x^5 + 1 = (x + 1) \left(x^2 - 2 \cos \frac\pi5 \cdot x + 1\right)\left(x^2 - 2 \cos \frac{3\pi}5 \cdot x + 1\right) = (x + 1) (x^2 - \phi x + 1) (x^2 - \bar\phi x + 1) ,$$ where $\phi := \frac12(1 + \sqrt 5)$ is the Golden Ratio and $\bar\phi := \frac12(1 - \sqrt 5)$ is its conjugate. So, our partial fraction decomposition has the form $$\frac{x^3}{x^5 + 1} = \frac{A}{x + 1} + \frac{B x + C}{x^2 - \phi x + 1} + \frac{D x + E}{x^2 - \bar\phi x + 1} .$$ Solving the resulting $5 \times 5$ linear system gives $$A = -\frac15, \quad B = E = \frac15 \phi, \quad C = D = \frac15 \bar\phi .$$