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If we want to integrate a quotient of two high degree polynomials like $$ \int\frac{x^a}{1+x^b}dx ~~~\text{with $a$, $b$ }\in\mathbb{N} $$ then I came to the conclusion in an exercise that if $0 < a = b - 1$, we get that $$\int\frac{x^a}{1+x^b}dx = \frac{\ln(1+x^b)}{b} + C$$ from integrating.

I wonder how to solve this if we don't have this specific relationship between $a$ and $b$. I'm not completely certain about this, but I imagine you can do integration by parts, integrating the $1+x^b$ part and the resulting $\ln$-terms until $x^a$ becomes $1$ (probably after $a$ times?). That method, however, seems very tedious and prone to making mistakes.

Is there good way of solving this when $a$ and $b$ are big but don't have that specific relationship?

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    $\begingroup$ Table of integrals series and products by I.S. Gradshteyn and I.M. Ryzhik (2007) Describes how to get a solution - essentially what you described (integration by parts) a and b integer. $\endgroup$ May 27, 2018 at 21:18
  • $\begingroup$ Do they mention if it can be done more efficiently with some sort of variable substitution or some other trick? @herbsteinberg $\endgroup$
    – dekuShrub
    May 27, 2018 at 21:24

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For the most general case, $$\int\frac{x^a}{1+x^b}\,dx=\frac{x^{a+1} }{a+1}\, _2F_1\left(1,\frac{a+1}{b};\frac{a+b+1}{b};-x^b\right)$$ where $_2F_1$ is the Gaussian or ordinary hypergeometric function.

Even if $a,b\in\mathbb{N}$, only a few of them have explicit expressions. If $a \geq b$, you can reduce the problem by long division and arrive to a polynomial plus this integral. You already found the case for $a=b-1$.

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