QUOTE(entryman @ Jan 2 2011, 06:40 PM)
Oh alright! So it's called integration by parts. I wonder why it's not in my book 
Found this: http://www.math.hmc.edu/calculus/tutorials/int_by_parts/
Pretty close example to my question, but still don't understand..
Gotta try to learn it up later............
Added on January 2, 2011, 6:41 pmThanks tanjiinjack
When you saw an integral in a form of u(x).v'(x), sometimes you'll need to use integration by parts, sometimes substitution.Found this: http://www.math.hmc.edu/calculus/tutorials/int_by_parts/
Pretty close example to my question, but still don't understand..
Gotta try to learn it up later............
Added on January 2, 2011, 6:41 pmThanks tanjiinjack
There are patterns for you to recognise. If u(x) is reducible, in a way that you can ultimately eliminate the x off by keep on differentiating, while v'(x) couldn't no matter how many times you differentiate (like e^x, sin x, cos x), then by parts will prove to be a very direct method to solve it.
So, in your problem, you need to choose 2x as your u(x), and the exponential stuff as your v'(x).
Write out your u(x) and v'(x), then derive your u'(x) and v(x).
Now, you should have 4 things, u(x), v(x), u'(x), v'(x).
Applying the formula: integral of (uv')= uv - integral of (u'v).
Then you should be able to solve it.
Note in the link you cite, it uses dv, while I use v'. They are same actually.
A bit extra, which might not be applicable.
» Click to show Spoiler - click again to hide... «
Jan 2 2011, 08:21 PM

Quote
0.0256sec
0.39
7 queries
GZIP Disabled