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?

There is nothing simple about this. My take would be to write
$$
\frac{x+1}{x^2-x+8}=\frac{x+1}{((x-1/2)^2+31/4)^3}
=\frac12\,\frac{2(x-1/2)+3}{((x-1/2)^2+31/4)^3}\\
=\frac12\,\frac{2(x-1/2)}{((x-1/2)^2+31/4)^3}
+
\frac12\,\frac{3}{((x-1/2)^2+31/4)^3}
$$
Now the first term is a straightforward substitution. The second one would require a couple rounds of integration by parts. 
According to Wolfram Alpha, an antiderivative is given by
$$
\int \frac{x+1}{(x^2-x+8)^3}\, dx = 
\frac{\frac{(31 (18 x^3-27 x^2+246 x-599)}{(x^2-x+8)^2}+36 \sqrt{31} \arctan\left(\frac{2 x-1}{\sqrt{31}}\right)}{59582}+c.
$$
So you cannot expect anything too straightforward. 
