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

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May 18th 2006, 08:41 AM
orange890

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

1
---------- dy
y^2 + 3y

Please help!!
May 18th 2006, 09:14 AM
CaptainBlack

Quote:

Originally Posted by orange890

1
---------- dy
y^2 + 3y

Please help!!

How do you find:

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

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

$ \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:

$ \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:

$ \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:

$ \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:

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

So it does :mad:

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