# Thread: An integral, partial fractions

1. ## An integral, partial fractions

This one is a really tough one for me. I'm betting an integration via partial fractions : $\displaystyle \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx$ with $\displaystyle a \neq 0$.
I could reach that the integral is worth $\displaystyle \int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx$. 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!

2. Originally Posted by arbolis
This one is a really tough one for me. I'm betting an integration via partial fractions : $\displaystyle \int \frac{2a^2x-6a+1}{a^2x^2-6ax+10}dx$ with $\displaystyle a \neq 0$.
I could reach that the integral is worth $\displaystyle \int 1+\frac{-a^2x^2+6ax-9+2a^2x-6a}{a^2x^2-6ax+10}dx$. 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:

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

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

3. Here's a hint for the second integral:

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

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

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

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

--Chris

4. Originally Posted by Chris L T521
Try this:

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

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

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

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

So firstly let

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

So

$\displaystyle dx=dz$

So we have

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

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

so

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

So we have

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

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

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

Backsubbing we get

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

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

So

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

5. Originally Posted by Mathstud28
$\displaystyle \int\frac{dx}{a^2x^2-6ax+10}$

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

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

So firstly let

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

So

$\displaystyle dx=dz$

So we have

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

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

so

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

So we have

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

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

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

Backsubbing we get

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

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

So

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

You made it too hard...

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

--Chris

6. Originally Posted by Chris L T521
You made it too hard...

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

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

7. 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

8. 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.