Evaluate the Integral

**Duffman** · February 16th 2009, 11:16 AM

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

Evaluate:

**Krizalid** · February 16th 2009, 11:34 AM

Okay, let's do some of algebra:

$\begin{aligned}<br /> \frac{x-5}{(x-4)(x-5)(x+5)}&=\frac{x+5-10}{(x-4)(x-5)(x+5)} \\ <br /> & =\frac{1}{(x-4)(x-5)}-\frac{10}{(x-4)(x-5)(x+5)} \\ <br /> & =\frac{1}{(x-4)(x-5)}-\underbrace{\frac{5}{4}\cdot \frac{(x-5)+(x+5)-2(x-4)}{(x-4)(x-5)(x+5)}}_{\alpha},<br /> \end{aligned}$

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!

**Jhevon** · February 16th 2009, 11:36 AM

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)

**Krizalid** · February 16th 2009, 11:38 AM

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

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