# Help with some integrals

• Apr 30th 2011, 11:01 AM
battleman13
Help with some integrals
Well I can't get the math latex to do what I want so here it is in laymans terms.

I am trying to help a friend with some what seems like basic calc stuff only it's been a while since I've done calc 1 and 2...

So I have several questions which I'll break down into three different new topics..

I am trying to evaluate the indefinite integral of 2X/X+3 dx

I put X+3 in the numerator like so (X+1)^ -1 but I'm not sure what to do from here... or if thats right.

Second question:

The definite integral evaluated from 7 to 10 of the absolute value of (X^2 - 81)/(X+9) dx

I realize that factors but would you do that? It would leave you with ((X+9)(X-9))/(X+9) which ultimate just leaves one factor of X-9 in the numerator.

So it would be the definite integral evaluated from 7 to 10 of absolute val (X-9)...

Any help would be greatly appreciated!
• Apr 30th 2011, 11:06 AM
TheEmptySet
Quote:

Originally Posted by battleman13
Well I can't get the math latex to do what I want so here it is in laymans terms.

I am trying to help a friend with some what seems like basic calc stuff only it's been a while since I've done calc 1 and 2...

So I have several questions which I'll break down into three different new topics..

I am trying to evaluate the indefinite integral of 2X/X+3 dx

I put X+3 in the numerator like so (X+1)^ -1 but I'm not sure what to do from here... or if thats right.

Second question:

The definite integral evaluated from 7 to 10 of the absolute value of (X^2 - 81)/(X+9) dx

I realize that factors but would you do that? It would leave you with ((X+9)(X-9))/(X+9) which ultimate just leaves one factor of X-9 in the numerator.

So it would be the definite integral evaluated from 7 to 10 of absolute val (X-9)...

Any help would be greatly appreciated!

For the first one you either need to do long division or algebraic manipulations to get

$\frac{2x}{x+3}=2\frac{x}{x+3}=2\frac{(x+3)-3}{x+3}=2\frac{x+3}{x+3}+2\frac{-3}{x+3}=2-\frac{3}{x+3}$

As I said you can also use polynomial long division.

You are correct on the 2nd one just be careful about where $|x-9|$ changes signs
• Apr 30th 2011, 11:19 AM
battleman13
Thanks! That helps a lot