# Integration Troubles

• Feb 25th 2009, 08:26 PM
Talwar
Integration Troubles
I'm working on a Linear Coefficients problem of the form G(ax+by):

$(3x-2y-6)dx-(2y-3x+2)dy=0$

I've managed to reduce it to the following,

$\int dx\ = \int -dz\frac{2-z}{5z-18}\\$

where $z = 3x-2y$

But I don't know how to finish this. I've scanned the table integrals in my book to no avail and I'm not sure how I'd use u substitution. Please lead me down the right path.
• Feb 25th 2009, 09:31 PM
matheagle
Divide, it's basic partial fractions.
You'll get $1+{something\over 5z-18}$

BUT there's something wrong, where is dy?
• Feb 25th 2009, 10:07 PM
Talwar
Sorry. I left it out. It's fix'd now.

As for the partial fractions, I think I figured it out. I used u-sub:

$u = 5z-18$

$du = 5dz$

$z = {u\over z}+{18\over 5}$

$dz = {du\over 5}$

And as long as I don't screw up the algebra, it should work. Thanks. ^_^
• Feb 25th 2009, 10:12 PM
matheagle
Quote:

Originally Posted by Talwar
Sorry. I left it out. It's fix'd now.

As for the partial fractions, I think I figured it out. I used u-sub:

$u = 5z-18$

$du = 5dz$

$z = {u\over z}+{18\over 5}$

$dz = {du\over 5}$

And as long as I don't screw up the algebra, it should work. Thanks. ^_^

What is $z = {u\over z}+{18\over 5}$ supposed to be?