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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?
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 $$\int \sqrt[a]{\tan t} \,dt .$$
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.
