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I have the following integral

$\int_a^b\frac{x^p}{C+x^q}\textrm{dx}$, where

  • C is a non-zero constant,
  • $a>0, b>0$, and
  • $p,q$ are rationals.

Is there a general solution for such an integral. I've found a related integral on Wikipedia: https://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_(a_+_b_xn)p

But there the exponents are integer and/or fractions. Any reference would be of great help.

Thanks.

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3 Answers 3

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For the antiderivative, there is a general solution for any $(p,q)$ $(c\neq 0)$which write $$\int\frac{x^p}{c+x^q}dx=\frac{x^{p+1} }{c (p+1)} \,_2F_1\left(1,\frac{p+1}{q};\frac{p+1}{q}+1;-\frac{x^q}{c}\right)$$ where appears the gaussian hypergeometric function.

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  • $\begingroup$ Thanks for your answer Claude. Can I ask for a reference where I could find these integrals? $\endgroup$ Sep 24, 2019 at 6:03
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Let $x^q=t$,Then $$\int \frac{x^p}{c+x^q} dx= \frac{1}{q}\int \frac {t^{p/q+1/q-1}}{c+t} dt= \frac{t^{1+r}}{cq(1+r)} ~2F_1[1,~1+r,~2+r;~\frac{-t}{c}], ~~~r=p/q+1/q-1,$$ $$ \Rightarrow I=\frac{x^{1+p}}{c(1+p)} ~_2F_1[1,~1+r,~2+r,-\frac{x^q}{c}]$$ where the hyper geometric funtion $~_2F_1$ is expressed as $$~_2F_1[A,B,C;z]=1+\frac{A B}{C} \frac{z}{1!}+\frac{A(A+1)B(B+1)}{C(C+1)}\frac{z^2}{2!}+...$$

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Let $c<0$ and $t=|c|x^q$, or $x=\sqrt[q]{\dfrac t{|c|}}$. The integral becomes

$$\int\frac{x^p}{c+x^q}dx=-|c|^{-(p+q+1)/q}\int\frac{t^{p/q}}{1- t}dx,$$

which is of the incomplete Beta type.

When $c>0$, a further change of variable $t:=\dfrac u{1-u}$ will turn it to the Beta from.

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