Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\,  dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more complicated then. 
Is there a way to solve this using only basic formulas?

As a last resort, it amounts to being able to compute
$$I_n=\int\frac{\mathrm d\mkern 1mu x}{(x^2-x+8)^n}$$
This sort of integrals can be calculated recursively. I'll show how to obtain a relation between $I_2$ and $I_3$. $I_1$ is standard, and is an $\arctan$ after a change of variable. 
Integrate $I_2$ by parts, setting
\begin{align*}u&=\frac1{(x^2-x+8)^2},&\mathrm d\mkern 1mu v&=\mathrm d\mkern 1mu x\\
\text{whence}\qquad \mathrm d\mkern 1mu u&=\frac{-2(2x-1)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x,& v&=x
\end{align*}
This yields
\begin{align*}I_2&=\frac{x}{(x^2-x+8)^2}+2\int\frac{(2x^2-x)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
&=\frac{x}{(x^2-x+8)^2}+2\int\frac{(2x^2-2x+16)+(x-8)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
&=\frac{x}{(x^2-x+8)^2}+4I_2+2\int\frac{x-8}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
\end{align*}
Now
$$\int\frac{x-8}{(x^2-x+8)^3}\mathrm d\mkern 1mu x=\int\frac{\frac12(2x-1)-\frac{15}2}{(x^2-x+8)^3}\mathrm d\mkern 1mu x=-\frac1{4(x^2-x+8)^2}-\frac{15}2I_3$$
from which we deduce the relation
$$15I_3-3I_2=\frac{2x-1}{(x^2-x+8)^2}.$$
Similarly, you can integrate $I_1$ by parts to obtain a relation between $I_2$ and $I_1$.
