1. ## Integrate function

hi
how do i integrate the following (what is the general format etc):

1/((3x+1)^2)

2. Originally Posted by taurus
hi
how do i integrate the following (what is the general format etc):

1/((3x+1)^2)
$\displaystyle \int \frac{dx}{(3x + 1)^2}$

Let $\displaystyle u = 3x + 1 \implies du = 3dx$, so
$\displaystyle \int \frac{dx}{(3x + 1)^2} = \frac{1}{3} \int \frac{du}{u^2}$

You can do it from there.

-Dan

3. does this seem right"

=> 1/2 (U^-1/-1)
=> -1/(2U)
?

4. Originally Posted by taurus
does this seem right"

=> 1/2 (U^-1/-1)
=> -1/(2U)
?
it's 1/3 not 1/2

5. ooop yea thanks

6. Originally Posted by taurus
ooop yea thanks
and of course, you should remember to replace u with 3x + 1 when done