Integration with partial fractions

**DBA** · February 22nd 2010, 11:11 AM

How do I start with this one?
$<br /> \int \frac {5x^{2}-3} {x^2-9}<br />$

Since nominator and denominator are x^2 I have to divide them, right?
I am not sure how to do that.

Thanks for any help.

**Ted** · February 22nd 2010, 11:14 AM

Originally Posted by DBA

How do I start with this one?
$<br /> \int \frac {5x^{2}-3} {x^2-9}<br />$

Since nominator and denominator are x^2 I have to divide them, right?
I am not sure how to do that.

Thanks for any help.

Yes start by Long Division.
Actually, there is another solutions without devide.

**DBA** · February 22nd 2010, 11:21 AM

Hello, thanks for answering. Can you please explain how to do so?
That was my problem, actually...

**Ted** · February 22nd 2010, 11:38 AM

Originally Posted by DBA

Hello, thanks for answering. Can you please explain how to do so?
That was my problem, actually...

Ohhh..You do not know to do the long division ?

**icemanfan** · February 22nd 2010, 11:44 AM

You could do it this way:

$\int \frac {5x^{2}-3} {x^2-9} dx =$

$\int \frac {5x^{2}-45 + 42} {x^2-9} dx =$

$\int \frac {5(x^2 - 9)}{x^2 - 9} dx + \int \frac {42}{x^2 - 9} dx=$

$\int 5 dx + \int \frac {42}{x^2 - 9} dx$

Then use an appropriate substitution on the second integral.

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