Answer
$$\int\frac{1}{x^6(x^5+4)}dx=\frac{1}{80}\ln|\frac{x^5+4}{x^5}|-\frac{1}{20x^5}+C$$
Work Step by Step
$$I=\int\frac{1}{x^6(x^5+4)}dx$$
Multiply both numerator and denominator with $x^4$:
$$I=\int\frac{x^4}{x^{10}(x^5+4)}dx$$
We set $u=x^5$, so $$du=5x^4dx$$ $$x^4dx=\frac{1}{5}du$$
Also, $x^{10}=(x^5)^2=u^2$
That means
$$I=\frac{1}{5}\int\frac{1}{u^2(u+4)}du$$
We will solve this integration using partial fractions method.
$$\frac{1}{u^2(u+4)}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u+4}$$
Clear the fractions: $$Au(u+4)+B(u+4)+Cu^2=1$$ $$Au^2+4Au+Bu+4B+Cu^2=1$$ $$(A+C)u^2+(4A+B)u+4B=1$$
We have
$4B=1$, so $B=1/4$
$4A+B=0$, so $A=-B/4=-1/16$
$A+C=0$, so $C=-A=1/16$
Therefore, $$I=\frac{1}{5}\Big(-\frac{1}{16}\int\frac{du}{u}+\frac{1}{4}\int\frac{du}{u^2}+\frac{1}{16}\int\frac{du}{u+4}\Big)$$ $$I=-\frac{1}{80}\int\frac{du}{u}+\frac{1}{20}\int\frac{du}{u^2}+\frac{1}{80}\int\frac{du}{u+4}$$ $$I=-\frac{1}{80}\ln|u|+\frac{1}{20}\Big(-\frac{1}{u}\Big)+\frac{1}{80}\ln|u+4|+C$$ $$I=\frac{1}{80}\ln|\frac{u+4}{u}|-\frac{1}{20u}+C$$ $$I=\frac{1}{80}\ln|\frac{x^5+4}{x^5}|-\frac{1}{20x^5}+C$$
