1. ## Definite Integral

$\int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$
break it up $6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)$
$2\int (e^udu-\ln{x}+c$

is this right so far, not sure what to do after.

2. Originally Posted by Jim Marnell
$\int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$
break it up $6\int_{x=1}^{x=2} (e^{3x}dx)-\int_{x=1}^{x=2} (\frac{1}{x}dx)$
$2\int (e^udu-\ln{x}+c$

is this right so far, not sure what to do after.
Well, yes, but this is a definite integral. So after you find the antiderivative $F(x),$ you need to apply the fundamental theorem of calculus by taking $F(2)-F(1).$

As for finding the antiderivative, what is $\int e^x\,dx?$ This is a basic integration rule, and replacing the $x$ with a $u$ won't make any difference.

3. yea i dont know why i put u in for the x, but, so now i would just insert the 2
(x) values in and subtract them

4. Originally Posted by Jim Marnell
yea i dont know why i put u in for the x, but, so now i would just insert the 2
(x) values in and subtract them
No, no, your substitution was fine. I was just pointing out the integration rule:

$\int e^x\,dx=e^x+C.$

So,

$\int_1^2\left(6e^{3x}-\frac1x\right)dx$

$=6\int_1^2e^{3x}\,dx-\int_1^2\frac1x\,dx$

$=2\int_{x=1}^{x=2}e^u\,du-\bigg[\ln\lvert x\rvert\bigg]_1^2$

$=2\bigg[e^u\bigg]_{x=1}^{x=2}-\ln 2$

$=2\bigg[e^{3x}\bigg]_1^2-\ln 2$

$=2(e^6-e^3)-\ln 2$

$=2e^3(e^3-1)-\ln 2$

5. Thanks!, thats what i ended up getting so hopefully i understand this concept now