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How to integrate the following function? $$\int \frac{x^{a}}{x^{2a}+k^{2}} dx$$ Where, $a>1$ and $k$ are constants.

Note: I tried doing it by doing partial fractions expansion on the denominator but couldn't proceed anywhere helpful. Is there a closed-form answer for this integration?

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  • $\begingroup$ See wolframalpha.com/… $\endgroup$ Jun 23 at 10:16
  • $\begingroup$ @SouparnaPal Thanks, but I actually came here to ask right after checking the wolfram alpha result, needed to know the procedure on how to,actually go through this integral. $\endgroup$ Jun 23 at 10:23
  • $\begingroup$ The point being made is that it's just a special function. Special function integrals are just plug and chug into the formula, there is no procedure. Look up the definitions of the functions involved on your own and try it. $\endgroup$ Jun 23 at 10:41
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    $\begingroup$ For integer $a$, see math.stackexchange.com/questions/3886239/…. For rational $a$, making an appropriate substitution $x = u^b$ to clear denominators in powers of $x$ transforms this integral to a rational one, which can be handled in the usual way. This integral arises naturally when evaluating $\int \sqrt[a]{\tan t}\,dt$; see math.stackexchange.com/questions/784591/… $\endgroup$ Jun 23 at 22:44

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When $k = 0$, the integrand is the power function $x^{-a}$, so we may as well take $k > 0$, in which case the substituting $x = \sqrt[a]{k} \,u$ transforms the integral to a constant multiple of $$\int \frac{u^a \,du}{u^{2 a} + 1} .$$

Now, if $a$ is a positive integer, this solution by Quanto gives the value \begin{multline*}-\frac1{4n} \sum_{k=1}^{2n} \Bigg[\cos\frac{(2k-1)(n + 1)\pi}{2n} \log\left(u^2-2u\cos \frac{(2k-1)\pi}{2n}+1\right)\\+2\sin\frac{(2k-1) (n + 1)\pi}{2n} \arctan\frac{\sin \frac{(2k-1)\pi}{2n}}{u-\cos \frac{(2k-1)\pi}{2n}} \Bigg].\end{multline*}

If $a$ is a rational number, say, $a = \frac{p}{q}$, then substituting $u = v^q$ transforms the integral to a rational one in $v$, which can then be handled using the usual methods, though even for particular $q > 1$ a formula for general $p$ in terms of elementary functions will be messy.

For general $a$, the integral in $u$ has value $$\frac{u^{a + 1}}{a + 1} {}_2F_1 \left(1, \frac{a + 1}{2a}; \frac{3 a + 1}{2a}; -u^{2 a} \right) + C ,$$ where ${}_2F_1$ is the ordinary hypergeometric function. It can also be written in terms of the Lerch transcendent $\Phi$ as $$\frac{u^{a + 1}}{2 a} \Phi\left(-u^{2a}, 1, \frac{a + 1}{2 a}\right) + C.$$

Remark (For $a > 1$) a standard contour integration gives the particular value $$\int_0^\infty \frac{u^a \,du}{u^{2 a} + 1} = \frac{\pi}{2 a} \sec \frac{\pi}{2 a} .$$

Remark The integral in $u$ arises in the evaluation of the integral $$\int \sqrt[a]{\tan t} \,dt .$$

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When $a$ is an integer, you can break the integrand into partial fractions, whose antiderivatives are given by a bunch of $\log$ and $\arctan$. (You could use this integral calculator to check the steps)

When $a$ is not an integer, there is in general no elementary function for your integral. But if you accept non-elementary solutions, then the integral can be expressed by hypergeometric functions.

That is, for any $a\geq0$ and $b,c>0$ we have the integral $$\int\frac{x^a}{x^b+c}dx=K+\frac{x^{1+a}}{c(1+a)}{_2F_1}\left(1,\frac{1+a}b;\frac{1+a}b+1;-\frac{x^b}c\right)$$ for any constant $K$, where $_2F_1$ is the hypergeometric function defined by

$$_2F_1(1,\beta;\beta+1;x)=\beta\int_0^1\frac{t^{\beta-1}}{\,1-xt\,}dt.$$


Proof: For any constant $K$, we have $$\int\frac{x^a}{x^b+c}dx=K+\frac1c\int_0^x\frac{u^a}{1+u^b/c}du.$$ Take $u=t^{1/b}x$, then $du=(x/b)t^{1/b-1}dt$ and it follows $$\frac1c\int_0^x\frac{u^a}{1+u^b/c}du=\frac{x^{1+a}}{bc}\int_0^1\frac{t^{\frac{1+a}b-1}}{\,1+(x^b/c)t\,}dt=\frac{x^{1+a}}{c(1+a)}{_2F_1}\left(1,\frac{1+a}b;\frac{1+a}b+1;-\frac{x^b}c\right).$$


Here are some short notes about hypergeometric functions if you are interested.

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