1. ## Evaluate the Integral

Here is another integral I’m having trouble with...

Evaluate:

2. Okay, let's do some of algebra:

\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 $\alpha$ equals $-\frac{5}{4}\left\{ \frac{1}{(x-4)(x+5)}+\frac{1}{(x-4)(x-5)}-\frac{2}{(x-5)(x+5)} \right\},$ and the integrand equals

$\frac{1}{(x-4)(x-5)}-\frac{5}{4}\cdot \frac{1}{(x-4)(x+5)}-\frac{5}{4}\left\{ \frac{1}{(x-4)(x-5)}-\frac{2}{(x-5)(x+5)} \right\},$ which can be rewritten as $-\frac{1}{4}\frac{1}{(x-4)(x-5)}-\frac{5}{4}\left\{ \frac{1}{(x-4)(x+5)}-\frac{2}{(x-5)(x+5)} \right\}.$

Now, I'll you show how to ged rid of the remaining partial fraction descomposition, so we have $\frac{1}{(x-4)(x-5)}=\frac{(x-4)-(x-5)}{(x-4)(x-5)}=\frac{1}{x-5}-\frac{1}{x-4}$ and you can do the same for the others. The rest is up to you and we're done!

3. Originally Posted by Duffman
Here is another integral I’m having trouble with...

Evaluate:
Note that $\int \frac {x - 5}{(x - 4)(x - 5)(x + 5)}~dx = \int \frac 1{(x - 4)(x + 5)}~dx = \frac 19 \int \left( \frac 1{x - 4} - \frac 1{x + 5} \right)~dx$ (via partial fractions or algebraic manipulaiton)

4. Ahahaha, I just now realized that, haha, too brute force for me!!