Finding the integral of $\frac{1}{x(x^n+a^n)}$ I was working on the problems in Mathematical Methods for Physics and Engineering by Riley,Hobson & Bence.
In Problem 2.34 (d) I'm supposed to find this integral: $$J=\int\frac{dx}{x(x^n+a^n)}.$$
I used partial fractions and arrived at the form
$$J=\frac{1}{a^n}\left[\log x-\int \frac{dx}{x^n+a^n}\right]$$
and now I'm stuck, I don't know how to integrate $1/(x^n+a^n)$.
 A: \begin{align*}
J&=\int\frac{dx}{x(x^n+a^n)}\\
&=\frac1{a^n}\int\left(\frac1{x}-\frac{x^{n-1}}{x^n+a^n}\right)dx\\
&=\frac1{a^n}\ln|x|-\frac1{na^n}\int\frac{nx^{n-1}}{x^n+a^n}dx\\
&=\frac1{a^n}\ln|x|-\frac1{na^n}\ln|x^n+a^n|+C,\\
\end{align*}
where $C$ is the constant of integration.
Alternatively, we have
\begin{align*}
J&=\int\frac{x^{-n-1}dx}{1+a^nx^{-n}}\\
&=-\frac1n\int\frac{(x^{-n})'dx}{1+a^nx^{-n}}\\
&=-\frac1{na^n}\int\frac{(1+a^nx^{-n})'dx}{1+a^nx^{-n}}\\
&=-\frac1{na^n}\ln|1+a^nx^{-n}|+C.\\
\end{align*}
A: Alternatively, perform a u-sub. Let $x=1/t$. What will happen is that
(after little algebraic simplification) you get a monomial numerator which is one degree lower than a binomial denominator consisting of an $x$ term and a constant. Integration: A basic $ln$ term. It's very easy. Try it out...
A: We seek a decomposition of the form
$$\frac{1}{x(x^n + a^n)} = \frac{A}{x} + \frac{Bx^{n-1}}{x^n + a^n} = \frac{(A+B)x^n + Aa^n}{x(x^n+a^n)}.$$  Hence the choice $A = a^{-n}$, $B = -A = -a^{-n}$, yields
$$\frac{1}{x(x^n+a^n)} = \frac{1}{a^n} \left(\frac{1}{x} - \frac{x^{n-1}}{x^n + a^n}\right).$$
The rest is straightforward:
$$\int \frac{dx}{x(x^n+a^n)} = \frac{1}{a^n} \left(\log |x| - \frac{1}{n} \int \frac{nx^{n-1}}{x^n + a^n} \, dx \right) = \frac{1}{a^n} \left( \log |x| -\frac{1}{n} \log |x^n + a^n| \right) + C.$$
A: \begin{gather*}
Let\ I=\int \frac{dx}{x\left( x^{n} +a^{n}\right)} =\int \frac{dx}{x^{n+1}} \cdotp \frac{1}{1+\frac{a^{n}}{x^{n}}}\\
Let\ 1+\frac{a^{n}}{x^{n}} =t\\
\frac{-n\cdotp a^{n}}{x^{n+1}} dx=dt\\
\frac{dx}{x^{n+1}} =\frac{-dt}{n\cdotp a^{n}}\\
I=\int \frac{-dt}{n\cdotp a^{n}} \cdotp \frac{1}{t} =\frac{-1}{n\cdotp a^{n}}\ln t=\frac{-1}{n\cdotp a^{n}}\ln\left( 1+\frac{a^{n}}{x^{n}}\right)\\
=\frac{-1}{n\cdotp a^{n}}\ln\left(\frac{x^{n} +a^{n}}{x^{n}}\right) =\frac{1}{a^{n}}\left(\ln x-\frac{1}{n}\ln\left( x^{n} +y^{n}\right)\right)\\
\end{gather*}
Hope this helps!
