# Thread: Intergration Question (fairly basic)

1. ## Intergration Question (fairly basic)

$\displaystyle \int(e^x - 3)^2 dx$

I think it goes as following, here is my working.

$\displaystyle \int (e^{2x} - 6e^x + 9) dx$

$\displaystyle = \frac{e^{2x}}{2} - e^x + 9x$

Can anyone confirm or tell me where I've gone wrong? Thank you very much.

2. Hi.

Originally Posted by Peleus
$\displaystyle \int(e^x - 3)^2 dx$

I think it goes as following, here is my working.

$\displaystyle \int (e^{2x} - 6e^x + 9) dx$

$\displaystyle = \frac{e^{2x}}{2} - e^x + 9x$

Can anyone confirm or tell me where I've gone wrong? Thank you very much.
Yea, the work seems to be correct, but you made a mistake

$\displaystyle \int - 6e^x dx \not= -e^x$

But I guess you know that

$\displaystyle = \frac{e^{2x}}{2} - 6e^x + 9x + const$

Yours
Rapha

3. Yep, sorry missed that.

$\displaystyle \int 6e^x$

which turns into

$\displaystyle 6 \int \frac{1}{6}e^x$

$\displaystyle 6 * e^x$

$\displaystyle 6e^x$

Thanks, forgot to type in the const as well.

4. Originally Posted by Peleus
Yep, sorry missed that.

$\displaystyle \int 6e^x$

which turns into

$\displaystyle 6 \int \frac{1}{6}e^x$
No!!!!

Originally Posted by Peleus
$\displaystyle 6 * e^x$
Yes

Originally Posted by Peleus
$\displaystyle 6e^x$
Yes

Actually it is $\displaystyle - \int 6e^x dx$ = $\displaystyle -6*\int e^xdx = -6*e^x$

Originally Posted by Peleus
Thanks
You're welcome

5. I have absolutely no idea what I was thinking there, even the working doesn't work.

Yes, you're right and I understand how you got it

Cheers.