# Thread: Help on another Definite Integral problem

1. ## Help on another Definite Integral problem

I'm lost on this one, just need a starting point.

$\int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$

Thanks for any help!

2. Originally Posted by Jim Marnell
I'm lost on this one, just need a starting point.

$\int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$

Thanks for any help!
for $6e^{3x}$ use the sub u=3x

The 2nd one needs to be memorized. Here is the hint

$\frac{d}{dx}\ln(x)=\frac{1}{x}$ so

$\int \frac{1}{x}dx=...$

3. Originally Posted by Jim Marnell
I'm lost on this one, just need a starting point.

$\int_{x=1}^{x=2} (6e^{3x}-\frac{1}{x})dx$

Thanks for any help!
Just integrate term by term and factor out your constants. I'll get you started:

$6 \int^{2}_{1} e^{3x} dx - \int^{2}_{1} \frac{1}{x} dx$

* For the first one, do a u-substitution with u = 3x

4. yep, thanks for getting me started!

5. for the 1st term, would it equal, $u=3x,,,, du=3,,, then,,, 6e^u(3)?$
I think thats the term thats confusing me the most

6. Originally Posted by Jim Marnell
for the 1st term, would it equal, $u=3x,,,, du=3,,, then,,, 6e^u(3)?$
I think thats the term thats confusing me the most
Consider $\int 6e^{3x}~dx$

Let $u = 3x$

$\Rightarrow du = 3~dx$

$\Rightarrow \frac 13 ~du = dx$

$\int 6e^{{\color{red}3x}}~{\color{blue}dx} = \int 6e^{\color{red}u} \cdot {\color{blue}\frac 13~du}$

$= 2 \int e^u~du$