1

---------- dy

y^2 + 3y

Please help!!

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- May 18th 2006, 08:41 AMorange890How to integrate 1/(y^2+3y) dy?
1

---------- dy

y^2 + 3y

Please help!! - May 18th 2006, 09:14 AMCaptainBlackQuote:

Originally Posted by**orange890**

$\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 AMTD!Quote:

Originally Posted by**CaptainBlack**

$\displaystyle

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

$ - May 18th 2006, 11:11 AMCaptainBlackQuote:

Originally Posted by**TD!**

RonL - May 18th 2006, 11:13 AMTD!
Don't be angry, be happy. Making mistakes is a sign of being (only?) human :D

- May 18th 2006, 01:01 PMtopsquarkQuote:

Originally Posted by**TD!**

-Dan