Here is another integral I’m having trouble with...
Evaluate:
Okay, let's do some of algebra:
$\displaystyle \begin{aligned}
\frac{x-5}{(x-4)(x-5)(x+5)}&=\frac{x+5-10}{(x-4)(x-5)(x+5)} \\
& =\frac{1}{(x-4)(x-5)}-\frac{10}{(x-4)(x-5)(x+5)} \\
& =\frac{1}{(x-4)(x-5)}-\underbrace{\frac{5}{4}\cdot \frac{(x-5)+(x+5)-2(x-4)}{(x-4)(x-5)(x+5)}}_{\alpha},
\end{aligned}$
thus $\displaystyle \alpha$ equals $\displaystyle -\frac{5}{4}\left\{ \frac{1}{(x-4)(x+5)}+\frac{1}{(x-4)(x-5)}-\frac{2}{(x-5)(x+5)} \right\},$ and the integrand equals
$\displaystyle \frac{1}{(x-4)(x-5)}-\frac{5}{4}\cdot \frac{1}{(x-4)(x+5)}-\frac{5}{4}\left\{ \frac{1}{(x-4)(x-5)}-\frac{2}{(x-5)(x+5)} \right\},$ which can be rewritten as $\displaystyle -\frac{1}{4}\frac{1}{(x-4)(x-5)}-\frac{5}{4}\left\{ \frac{1}{(x-4)(x+5)}-\frac{2}{(x-5)(x+5)} \right\}.$
Now, I'll you show how to ged rid of the remaining partial fraction descomposition, so we have $\displaystyle \frac{1}{(x-4)(x-5)}=\frac{(x-4)-(x-5)}{(x-4)(x-5)}=\frac{1}{x-5}-\frac{1}{x-4}$ and you can do the same for the others. The rest is up to you and we're done!