# How to integrate 1/(y^2+3y) dy?

• May 18th 2006, 08:41 AM
orange890
How to integrate 1/(y^2+3y) dy?
1
---------- dy
y^2 + 3y

• May 18th 2006, 09:14 AM
CaptainBlack
Quote:

Originally Posted by orange890
1
---------- dy
y^2 + 3y

How do you find:

$\displaystyle \int \frac{1}{y^2+3y} \ dy$?

You resolve the integrand into partial fractions, which can then be integrated
by inspection.

$\displaystyle \frac{1}{y^2+3y}=\frac{1}{y(y+3)}=\frac{A}{y}+ \frac{B}{y+3}$

With a bit of manipulation you will find:

$\displaystyle \frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}- \frac{1}{y+3} \right\}$.

You should now be able to find the integral yourself from this point.

RonL
• May 18th 2006, 11:06 AM
TD!
Quote:

Originally Posted by CaptainBlack
With a bit of manipulation you will find:

$\displaystyle \frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{-1}{y}+ \frac{1}{y+3} \right\}$.

I believe the signs should be switched, so:

$\displaystyle \frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}+ \frac{-1}{y+3} \right\}$
• May 18th 2006, 11:11 AM
CaptainBlack
Quote:

Originally Posted by TD!
I believe the signs should be switched, so:

$\displaystyle \frac{1}{y^2+3y}=\frac{1}{3}\left\{ \frac{1}{y}+ \frac{-1}{y+3} \right\}$

RonL
• May 18th 2006, 11:13 AM
TD!
Don't be angry, be happy. Making mistakes is a sign of being (only?) human :D
• May 18th 2006, 01:01 PM
topsquark
Quote:

Originally Posted by TD!
Don't be angry, be happy. Making mistakes is a sign of being (only?) human :D

Hey, I once had a Mechanics prof who would take 15 points off for missing a "-" sign. (Yes, it WAS out of a 100 point exam.) The bad news was that his lectures contained just those sorts of errors. (We were all happy to get B's in that class!)

-Dan