# intergration

• Jun 10th 2007, 07:44 PM
startingover
intergration
I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit.

integrate 3/(2x+5)dx

=3(1/x^2+5x+C)
• Jun 10th 2007, 07:47 PM
ThePerfectHacker
Quote:

Originally Posted by startingover
I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit.

integrate 3/(2x+5)dx

Hint: $\int \frac{3}{2x+5} = 3\int \frac{1}{2x+5} dx$
• Jun 10th 2007, 07:48 PM
Jhevon
Quote:

Originally Posted by startingover
I am not sure if I am integrating this correctly. The constant in the numerator is throwing me off a bit.

integrate 3/(2x+5)dx

=3(1/x^2+5x+C)

$\int \frac {3}{2x + 5}dx = 3 \int \frac{1}{2x + 5} dx$

Now use substitution, $u = 2x + 5$

Can you take it from here?
• Jun 10th 2007, 07:56 PM
Krizalid
Quote:

Originally Posted by startingover
integrate 3/(2x+5)dx

$\int\frac3{2x+5}~dx=3\int\frac1{2x+5}~dx=\frac32\i nt\frac{(2x+5)'}{2x+5}~dx=\frac32\ln\left|2x+5\rig ht|+k$