Question

ingrat (6x-11)/((x-1)(x^2-4x+8)dx

Answer 1

Partial fraction decomp:
\[\begin{align*}\frac{6x-11}{(x-1)(x^2-4x+8)}&=\frac{A}{x-1}+\frac{Bx+C}{x^2-4x+8}\\
6x-11&=Ax^2-4Ax+8A+Bx^2+Cx-Bx-C\\
&=(A+B)x^2+(-4A-B+C)x+(8A-C)
\end{align*}\]
Matching up coefficients yields:
\[\begin{cases}A+B=0\\-4A-B+C=6\\8A-C=-11\end{cases}~~\Rightarrow~~\begin{cases}A=-1\\B=1\\C=3\end{cases}\]
So now you integrate:
\[-\int\frac{dx}{x-1}+\int\frac{x+3}{x^2-4x+8}~dx\]
The first is trivial. For the second, there are a few ways to go about it. I would suggest the following:
\[\int\frac{x+3}{x^2-4x+8}=\int\frac{x-2}{(x-2)^2+4}~dx+5\int\frac{dx}{(x-2)^2+4}\]
If you didn't follow, this was just a matter of completing the square in the denominator and rewriting and breaking up the numerator. Next, you'd substitute, letting \(u=x-2\) so that \(du=dx\):
\[-\int\frac{dx}{x-1}+\int\frac{u}{u^2+4}~du+5\int\frac{du}{u^2+4}\]
And finally, a trig substitution. I'll leave that to you.