How to find the integral $\int \frac{\sqrt{1+x^{2n}}\left(\log(1+x^{2n}) -2n \log x\right)}{x^{3n+1}}dx$? How to evaluate the integral : 
$$\int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}$$
I have attempted an evaluation, but I am at a loss as to a useful result. Thanks for any and all help.
 A: By using the change of variable $$u=\dfrac1{x^{2n}},\quad \log u = -2n \log x, \quad \dfrac{du}u=-2n\dfrac{dx}x,$$ then an integration by parts, one gets
$$
\begin{align}
&\int \frac{\sqrt{1+x^{2n}}\{\log(1+x^{2n}) -2n \log x\}}{x^{3n+1}}dx\\\\
&= \int \frac{\sqrt{1+x^{2n}}\:\log\left(1+\frac1{x^{2n}}\right)}{x^{3n+1}}dx\\\\
&=\int \frac1{x^{2n}}\sqrt{1+\frac1{x^{2n}}}\:\log\left(1+\frac1{x^{2n}}\right)\frac{dx}{x}\\\\
&=-\frac1{2n}\int u\sqrt{1+u}\:\log\left(1+u\right)\frac{du}{u}\\\\
&=-\frac1{2n}\int \sqrt{1+u}\:\log\left(1+u\right)du\\\\
&=-\frac1{3n} (1+u)^{3/2}\:\log\left(1+u\right)+\frac1{3n}\int \sqrt{1+u}\:du+C\\\\
&=-\frac1{3n} (1+u)^{3/2}\:\log\left(1+u\right)+\frac2{9n} (1+u)^{3/2}+C\\\\
&=-\frac1{3n} \left(1+\frac1{x^{2n}}\right)^{3/2}\:\log\left(1+\frac1{x^{2n}}\right)+\frac2{9n} \left(1+\frac1{x^{2n}}\right)^{3/2}+C.
\end{align}
$$
A: Let
\begin{align}
I = \int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}
\end{align}
and make the transformation $x = t^{1/2n}$ to obtain
\begin{align}
I = \frac{1}{2n} \, \int t^{-5/2} \, \sqrt{1+t} \, \left(\ln(1+t) - \ln(t)\right) \, dt.
\end{align}
Let $t = \sinh^{2}\theta$ to obtain
\begin{align}
\frac{n}{2} \, I &= \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\cosh\theta) \, d\theta - \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\sinh\theta) \, d\theta = J_{1} - J_{2}
\end{align}
\begin{align}
J_{1} &= \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\cosh\theta) \, d\theta \\
&= \int \ln(\cosh\theta) \, \partial_{\theta}\left(\frac{-1}{3} \, \coth^{3}\theta \right) \, d\theta \\
&= - \frac{1}{3} \, \coth^{3}\theta \, \ln(\cosh\theta) + \frac{1}{3} \, \int \coth^{2}\theta \, d\theta \\
&= \frac{1}{3} \left[ \theta - \coth\theta - \coth^{3}\theta \, \ln(\cosh\theta) \right]
\end{align}
In a similar manor
\begin{align}
J_{2} = \frac{1}{3} \left[ \theta - \frac{1}{3} \, \coth\theta \, (csch^{2}\theta + 4) - \coth^{3}\theta \, \ln(\sinh\theta) \right].
\end{align}
Now,
\begin{align}
I = \frac{2}{9 \, n} \left[ \coth\theta \, (1 + csch^{2}\theta) - \coth^{3}\theta \, \ln(\coth\theta) \right].
\end{align}
By reversing the substitutions the result
\begin{align}
I = \frac{2}{9n} \left(1+\frac{1}{x^{2n}}\right)^{3/2} - \frac{1}{3n} \left(1+\frac1{x^{2n}}\right)^{3/2} \, \ln\left(1+\frac1{x^{2n}}\right) + c_{0}
\end{align}
is obtained. 
