Author Topic: integration help (Read 1566 times) Tweet Share

sajib_mostofa · « **on:** October 25, 2010, 12:28:10 pm »

Find $\int e^{6x}\sqrt{e^{2x}+1}$

qwertyda2nd · « **Reply #1 on:** October 25, 2010, 12:40:24 pm »

page 9.

Linkage1992 · « **Reply #2 on:** October 25, 2010, 12:42:38 pm »

cool i got that answer and was just about to post it up.

m@tty · « **Reply #3 on:** October 25, 2010, 12:44:00 pm »

You're missing a $dx$ there...

$I=\int e^{6x}\sqrt{e^{2x}+1}$

let $u = e^{2x}+1$

$e^{2x}=(u-1)$

$\frac{du}{dx}=2e^{2x}$

$I=\frac{1}{2}\int e^{4x}\sqrt{e^{2x}+1}\cdot 2e^{2x}dx$

$=\frac{1}{2} \int \left(e^{2x}\right)^2 \sqrt{u}du$

$=\frac{1}{2} \int (u-1)^2\sqrt{u}du$

$=\frac{1}{2} \int u^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}} du$

$=\frac{1}{2} \left( \frac{2}{7}u^\frac{7}{2} -\frac{4}{5}u^\frac{5}{2}+\frac{2}{3}u^\frac{3}{2}\right) +C$

$=\frac{1}{7}(e^{2x}+1)^\frac{7}{2}-\frac{2}{5}(e^{2x}+1)^\frac{5}{2}+\frac{1}{3}(e^{2x}+1)^\frac{3}{2}+C$

EDIT: Beaten.. damn.

sajib_mostofa · « **Reply #4 on:** October 25, 2010, 12:49:34 pm »

I don't understand how you go from $e^{6x}$ to $e^{4x}$

98.40_for_sure · « **Reply #5 on:** October 25, 2010, 12:50:45 pm »

He 'used up' the other e^(2x) to make the du

sajib_mostofa · « **Reply #6 on:** October 25, 2010, 12:52:39 pm »

Quote from: 99.95_for_sure on October 25, 2010, 12:50:45 pm

He 'used up' the other e^(2x) to make the du

Ahh yep I see. Cheers for that.

m@tty · « **Reply #7 on:** October 25, 2010, 12:52:59 pm »

Notice that there is a $e^{2x}$ term at the end of the integrand.

$\frac{1}{2}\int \underbrace{\underbrace{e^{4x}}_{HERE} \sqrt{e^{2x}+1} \underbrace{2e^{2x}}_{HERE}}_{AND \ e^{4x}\cdot e^{2x}=e^{6x}} dx$

EDIT: Too slow. Again.

sajib_mostofa · « **Reply #8 on:** October 25, 2010, 01:11:36 pm »

Quote from: m@tty on October 25, 2010, 12:52:59 pm

Notice that there is a $e^{2x}$ term at the end of the integrand.

$\frac{1}{2}\int \underbrace{\underbrace{e^{4x}}_{HERE} \sqrt{e^{2x}+1} \underbrace{2e^{2x}}_{HERE}}_{AND \ e^{4x}\cdot e^{2x}=e^{6x}} dx$

EDIT: Too slow. Again.

Ah well you made it more clear to me

m@tty · « **Reply #9 on:** October 25, 2010, 01:14:38 pm »

Yeah, I thought it looked cool

98.40_for_sure · « **Reply #10 on:** October 25, 2010, 01:15:17 pm »

Quote from: m@tty on October 25, 2010, 01:14:38 pm

Yeah, I thought it looked cool

*showoff

jasoN- · « **Reply #11 on:** October 25, 2010, 01:15:52 pm »

nice Latex-ing skills there matty

m@tty · « **Reply #12 on:** October 25, 2010, 01:20:58 pm »

Quote from: 99.95_for_sure on October 25, 2010, 01:15:17 pm

Quote from: m@tty on October 25, 2010, 01:14:38 pm
Yeah, I thought it looked cool

*showoff

Touché.

Mao · « **Reply #13 on:** October 25, 2010, 01:46:41 pm »

Quote from: m@tty on October 25, 2010, 12:52:59 pm

Notice that there is a $e^{2x}$ term at the end of the integrand.

$\frac{1}{2}\int \underbrace{\underbrace{e^{4x}}_{HERE} \sqrt{e^{2x}+1} \underbrace{2e^{2x}}_{HERE}}_{AND \ e^{4x}\cdot e^{2x}=e^{6x}} dx$

EDIT: Too slow. Again.

I am impressed. Do you LaTeX your MUEP assignments?

m@tty · « **Reply #14 on:** October 25, 2010, 01:52:51 pm »

I did with a few, but they didn't turn out so well.

I haven't bothered figuring out how to use a proper program(I downloaded one I think you recommended in some thread, but couldn't get it working properly), so I just typed them up on here and copy-pasted.

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Author Topic: integration help (Read 1566 times) Tweet Share

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