Finding $\int\frac{1}{x^{11}+4x^6}dx$ I wanted to find out if there is an easy way to evaluate $$\displaystyle\int\frac{1}{x^{11}+4x^6}dx.$$
I substituted $u=x^5$ and then used partial fractions, but maybe there is a simpler way to find this.
 A: Hint:
$$
\begin{align}
\int\frac{dx}{x^{11}+4x^6}&=\int\frac{x^4dx}{x^{15}+4x^{10}}\\
&=\frac14\int\frac{du}{u^3+4u^2}\,\,\,\,\,,u=x^5\\
&=\frac{1}{16}\int\bigg[\frac{1}{u^2}-\frac{1}{4u}+\frac{1}{4(4+u)}\bigg]du
\end{align}
$$
A: After using partial fractions, try the substitution $u=x^5+4$ to deal with the first integral in the expression
\begin{equation*}
\frac{1}{16}\int\frac{x^4}{x^5+4}dx-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx
\end{equation*}
which is what you get when you use partial fractions on your integrand. 
A: Let $t=\frac{1}{x}$, so then $x=\frac{1}{t}$ and $dx=-\frac{1}{t^2}dt$.
Then $\displaystyle\int\frac{1}{x^{11}+4x^6}dx=\int\frac{1}{\frac{1}{t^{11}}+\frac{4}{t^6}}\left(-\frac{1}{t^2}\right)dt=-\int\frac{t^9}{1+4t^5}dt$
$=-\int\big(\frac{1}{4}t^4-\frac{\frac{1}{4}t^4}{1+4t^5}\big)dt=-\frac{1}{20}t^5+\frac{1}{80}\ln\vert1+4t^5\vert+C=-\frac{1}{20x^5}+\frac{1}{80}\ln\vert1+\frac{4}{x^5}\vert+C$
A: $$\int\frac{1}{x^{11}+4x^6}dx=$$
$$\int \left(\frac{1}{4x^6}+\frac{x^4}{16(x^5+4)}-\frac{1}{16x}\right)dx=$$
$$\frac{1}{16}\int \frac{x^4}{x^5+4}dx-\frac{1}{16}\int \frac{1}{x}dx+\frac{1}{4}\int \frac{1}{x^6}dx=$$
$$\frac{1}{80}\int \frac{1}{u}du-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{1}{16}\int\frac{1}{x}dx+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{\ln(x)}{16}+\frac{1}{4}\int\frac{1}{x^6}dx=$$
$$\frac{\ln(u)}{80}-\frac{1}{20x^5}-\frac{\ln(x)}{16}+C=$$
$$\frac{1}{80}\left(-\frac{4}{x^5}+\ln(x^5+4)-5\ln(x)\right)+C$$
