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?

We don't need to use any integration by parts.
We start by completing the square
$$\int \frac{x+1}{(x^2-x+8)^3}  dx = \int \frac{x-\frac{1}{2}+\frac{3}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx =$$
$$\underbrace{ \int \frac{x-\frac{1}{2}}{((x-\frac{1}{2})^2+\frac{31}{4})^3} dx}_{=:I_1}+ \frac{3}{2}\underbrace{\int\frac{1}{((x-\frac{1}{2})^2+\frac{31}{4})^3}dx}_{=:I_2}$$
For $I_1$ substitution $(x-\frac{1}{2})^2 = t$ then $\frac{dt}{dx}= 2(x-\frac{1}{2})$ so 
$$I_1 = \frac{1}{2}\int \frac{1}{t^3}dt = -\frac{1}{4t^2}+C$$
For $I_2$ first substitution $x-\frac{1}{2}=t$ 
$$I_2 = \int \frac{1}{(t^2+\frac{31}{4})^3}dt$$
then rearrange
$$ \int \frac{1}{(t^2+\frac{31}{4})^3}dt = \frac{4^3}{31^3}\int \frac{1}{{\left(1+\left(\frac{2t}{\sqrt{31}}\right)^2\right)}^3}dt.$$
Now another substitution $\frac{2t}{\sqrt{31}}=z$ gives us the following
$$\frac{4^3}{31^3}\int \frac{1}{{\left(1+\left(\frac{2t}{\sqrt{31}}\right)^2\right)}^3}dt = \left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{(1+z^2)^3}dz.$$
The last substitution $z=\tan{y}$ and $dz= \frac{1}{\cos^2y}dy$, and we get something much simpler
$$\left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{(1+z^2)^3}dz = \left(\frac{4}{31}\right)^\frac{5}{2}\int \frac{1}{\cos^2y}\cos^6(y)dy= \left(\frac{4}{31}\right)^\frac{5}{2}\int \cos^4(y)dy$$
Now we use that $\cos^2 y = {1 + \cos(2y) \over 2}$ to show that $\cos^4 y = \frac{\cos(4y) + 4\cos(2y)+3}{8}$ and thus 
$$\left(\frac{4}{31}\right)^\frac{5}{2}\int \cos^4(y)dy = \left(\frac{4}{31^\frac{5}{2}}\right)\int (\cos(4y) + 4\cos(2y)+3)dy$$
which is easy to find. Now you have to go backwards with all the substitutions to get the result with variable $x$ only. 
