An integral, partial fractions

**arbolis** · July 19th 2008, 07:04 PM

This one is a really tough one for me. I'm betting an integration via partial fractions : $\int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx$ with $a \neq 0$ .
I could reach that the integral is worth $\int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx$ . I don't know how to proceed from it. A first attempt would be to factorize that denominator, but it's not an easy task for me!

**Chris L T521** · July 19th 2008, 07:09 PM

Originally Posted by arbolis

This one is a really tough one for me. I'm betting an integration via partial fractions : $\int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx$ with $a \neq 0$ .
I could reach that the integral is worth $\int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx$ . I don't know how to proceed from it. A first attempt would be to factorize that denominator, but it's not an easy task for me!

Try this:

$\int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}\,dx=\int\frac{2a^2x-6a}{a^2x^2-6ax+10}\,dx+\int\frac{1}{a^2x^2-6ax+10}\,dx$

The first one can be solved using substitution.

The second one you'll have to experiment with. I'll work on it now. I'll post something when I get it.

--Chris

**Chris L T521** · July 19th 2008, 07:21 PM

Here's a hint for the second integral:

$\int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx$

$\implies\int\frac{1}{(ax-3)^2+1}\,dx$

This is something you can integrate. Take it from here!

This one didn't need partial fractions after all...

--Chris

**Mathstud28** · July 19th 2008, 07:29 PM

Originally Posted by Chris L T521

Try this:

$\int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}\,dx=\int\frac{2a^2x-6a}{a^2x^2-6ax+10}\,dx+\int\frac{1}{a^2x^2-6ax+10}\,dx$

The first one can be solved using substitution.

The second one you'll have to experiment with. I'll work on it now. I'll post something when I get it.

--Chris

$\int\frac{dx}{a^2x^2-6ax+10}$

$=\frac{1}{a^2}\frac{dx}{x^2-\frac{6}{a}x+\frac{10}{a^2}}$

$=\frac{1}{a^2}\int\frac{dx}{\left(x-\frac{3}{a}\right)^2+\frac{1}{a^2}}$

So firstly let

$z=x-\frac{3}{a}$

So

$dx=dz$

So we have

$\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}$

Now let $z=\frac{1}{a}\tan(\theta)$

so

$dz=\frac{1}{a}\sec^2(\theta)$

So we have

$\frac{1}{a^2}\int\frac{\frac{1}{a}\sec^2(\theta)}{ \frac{1}{a^2}\tan^2(\theta)+\frac{1}{a^2}}~d\theta$

$=\frac{1}{a}\int~d\theta$

$=\frac{\theta}{a}$

Backsubbing we get

$\int\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}$

$=\frac{1}{a}\arctan\left(za\right)$

So

$\int\frac{dx}{a^2x^2-6ax+10}=\frac{\arctan\left(ax-3\right)}{a}$

**Chris L T521** · July 19th 2008, 07:32 PM

Originally Posted by Mathstud28

$\int\frac{dx}{a^2x^2-6ax+10}$

$=\frac{1}{a^2}\frac{dx}{x^2-\frac{6}{a}x+\frac{10}{a^2}}$

$=\frac{1}{a^2}\int\frac{dx}{\left(x-\frac{3}{a}\right)^2+\frac{1}{a^2}}$

So firstly let

$z=x-\frac{3}{a}$

So

$dx=dz$

So we have

$\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}$

Now let $z=\frac{1}{a}\tan(\theta)$

so

$dz=\frac{1}{a}\sec^2(\theta)$

So we have

$\frac{1}{a^2}\int\frac{\frac{1}{a}\sec^2(\theta)}{ \frac{1}{a^2}\tan^2(\theta)+\frac{1}{a^2}}~d\theta$

$=\frac{1}{a}\int~d\theta$

$=\frac{\theta}{a}$

Backsubbing we get

$\int\frac{1}{a^2}\int\frac{dz}{z^2+\frac{1}{a^2}}$

$=\frac{1}{a}\arctan\left(za\right)$

So

$\int\frac{dx}{a^2x^2-6ax+10}=\frac{\arctan\left(ax-3\right)}{a}$

You made it too hard...

$\int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx=\int\frac{1}{(ax-3)^2+1}\,dx$ $=\frac{1}{a}\tan^{-1}(ax-3)+C$

--Chris

**Mathstud28** · July 19th 2008, 07:33 PM

Originally Posted by Chris L T521

You made it too hard...

$\int\frac{1}{a^2x^2-6ax+10}\,dx=\int\frac{1}{a^2x^2-6ax+9+1}\,dx=\int\frac{1}{(ax-3)^2+1}\,dx$ $=\frac{1}{a}\tan^{-1}(ax-3)+C$

--Chris

I didn't make anything too hard, I just did it in a different style than you.

**Chris L T521** · July 19th 2008, 07:35 PM

Originally Posted by Mathstud28

I didn't make anything too hard, I just did it in a different style than you.

Ok. Fine. It wasn't too hard. It was just too long...

Your technique is very interesting. I like it.

--Chris

**Mathstud28** · July 19th 2008, 07:39 PM

Originally Posted by Chris L T521

Ok. Fine. It wasn't too hard. It was just too long...

--Chris

I just showed ALL my steps.

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