integrals

[legacyofpiracy]

Hello, having some trouble with these two integrals:

S (2x+3)e^(4x) dx 

and 


S (dx)/(e^-2x)

would you recommend I use integration by parts in these two?

 

[Unco]

Why not give it a try, for the first one at least -- your second one looks dodgy.

 

[galactus]

Integration by parts:

Let \(\displaystyle u=2x+3\)

\(\displaystyle dv=e^{4x}dx\)

\(\displaystyle v=\frac{e^{4x}}{4}dx\)

\(\displaystyle du=2dx\)

Now put it together:

\(\displaystyle \int{udv}=uv-\int{vdu}\)

 

[legacyofpiracy]

alright thanks, so for this first one i have

(2x+3)(e^4x/4)- S (e^4x/4)(2dx)

will I need to go any further with this? Such as do another integration by parts?

Also what do you mean by dodgy unco?

 

[Unco]

What is the integral of e^(4x)/2 ?

And the same goes for the dodgy one.

 

[legacyofpiracy]

could you do e^4xln(4) or does that break a whole bunch of rules?

 

[Unco]

Lest we forget the basics.

\(\displaystyle \L\mbox{ \int e^{ax} dx = \frac{1}{a}e^{ax} + C}\)

 

[legacyofpiracy]

so (2x+3)(e^4x/4)-(1/16e^4x)(2x) ?

 

[Unco]

\(\displaystyle \L\mbox{ \int \frac{e^{4x}}{4} (2 dx) = \int \frac{e^{4x}}{2} dx = \frac{e^{4x}}{8} + C}\)

 

[legacyofpiracy]

ooh alright. so for the second one should I do the same thing and use integration by parts?

 

[Unco]

The second one is just \(\displaystyle \mbox{\int e^{2x} dx}\). You can integrate this straight off.

 

[legacyofpiracy]

oh! I always just have trouble seeing these sort of things. thank you