Integral problem--Integral (e^2x-2x)^2(e^2x-1) dx

[Linty Fresh]

OK, so I start out with this:

Integral (e^2x-2x)^2(e^2x-1) dx

And I want to use substitution to solve it. I know that the derivative of (e^2x-2x) is e^2x-2, but I'm not sure what to multiply (e^2x-1) dx by to get that? Could someone please help me get started? Or am I going about this wrong? Should I just multiply the whole thing out and find the integral of the total? Many thanks.

 

[royhaas]

Yes, just expand and integrate termwise.

 

[skeeter]

OK, so I start out with this:

Integral (e^2x-2x)^2(e^2x-1) dx

And I want to use substitution to solve it. I know that the derivative of (e^2x-2x) is e^2x-2, ...

no, it's 2e^2x - 2 = 2(e^2x - 1)

now you're set for using a substitution.

 

[soroban]

Hello, Linty Fresh!

\(\displaystyle \L\int \left(e^{2x}\,-\,2x\right)^2\left(e^{2x}\,-\,1)\, dx\)

Let: \(\displaystyle \,u \:=\:e^{2x}\,-\,2x\;\;\Rightarrow\;\;du \:=\:\left(2e^{2x}\,-\,2\right)\,dx\:=\:2\left(e^{2x}\,-\,1\right)\,dx\;\;\Rightarrow\;\;\left(e^{2x}\,-\,1\right)dx \:=\:\frac{1}{2}\,du\)

Substitute: \(\displaystyle \L\:\int\underbrace{\left(e^{2x}\,-\,2x\right)^2}\underbrace{\left(e^{2x}\,-\,1\right)\,dx}\)
. . . . . . . . . . . \(\displaystyle \L\;\int\;\;u^2\;\;\;\;\;\;\;\;\;\;\frac{1}{2}\,du\)

We have: \(\displaystyle \L\;\frac{1}{2}\int u^2\,du \;=\;\frac{1}{6}\,u^3\,+\,C\)

Back-substitute: \(\displaystyle \L\;\frac{1}{6}\left(e^{2x}\,-\,2x\right)^3\,+\,C\)

 

[Linty Fresh]

Just wanted to post to thank you all. I wanted to thank you before, but I was offline for a couple of days and then the board was down, so thanks again!