I wan't too sure where to put this. It is Calculus, but not University level. Any how,
I need to integrate:
1/(3x+7)
Now I'd be fine with it if the power wasn't -1 seeing as I get 0 on the bottom. I think it has to something do with lnx.
Thanks.
I wan't too sure where to put this. It is Calculus, but not University level. Any how,
I need to integrate:
1/(3x+7)
Now I'd be fine with it if the power wasn't -1 seeing as I get 0 on the bottom. I think it has to something do with lnx.
Thanks.
That's why it doesn't work..
$\displaystyle \int{\frac{1}{x}}dx=ln|x|$
$\displaystyle \int{\frac{1}{u}}du=ln|u|$
Hence $\displaystyle u=3x+7\ \Rightarrow\ \frac{du}{dx}=3\ \Rightarrow\ du=3dx\ \Rightarrow\ dx=\frac{du}{3}$
The integral is now solved using
$\displaystyle \frac{1}{3}\int{\frac{1}{u}}du=\frac{1}{3}ln|u|=\f rac{1}{3}ln|3x+7|$