Hi everyone!

I am having trouble doing this integration by parts. Mine got really ugly and it doesn't seem like is should. Can y'all work this for me?

$\displaystyle \int\limits_{1}^{8} 5\cdot\frac{(\ln{x})^2}{x^3} dx$

Thanks!

Molly

Printable View

- Feb 1st 2009, 05:52 PMmollymcf2009Integration by parts
Hi everyone!

I am having trouble doing this integration by parts. Mine got really ugly and it doesn't seem like is should. Can y'all work this for me?

$\displaystyle \int\limits_{1}^{8} 5\cdot\frac{(\ln{x})^2}{x^3} dx$

Thanks!

Molly - Feb 1st 2009, 07:05 PMKrizalid
Yo, Molly.

Read my signature about how to increase your speed when integrating by parts.

Now, if you want to make it to the classic way, put $\displaystyle u=\ln^2x$ and $\displaystyle dv=\frac{dx}{x^3}.$ (You're gonna have to make another integration by parts after this one.) - Feb 1st 2009, 07:28 PMmollymcf2009
WOW!! Thanks so much for that! (Clapping)

- Feb 1st 2009, 08:04 PMmollymcf2009
Using that method you showed me, do I have this right?

$\displaystyle \int\limits_{1}^{8} 5 \cdot (lnx)^2 \cdot (x^{-3})' dx$

$\displaystyle \frac{5}{2} \cdot (\frac{(lnx)^2}{x^2} + \frac{2lnx}{2x^3}) $ evaluated from 1 to 8 - Feb 2nd 2009, 05:20 AMskeeter
let $\displaystyle t = \ln{x}$

$\displaystyle x = e^t$

$\displaystyle

dt = \frac{1}{x} \, dx

$

$\displaystyle

\int \frac{(\ln{x})^2}{x^2} \cdot \frac{1}{x} \, dx

$

substitute ...

$\displaystyle

\int \frac{t^2}{e^{2t}} \, dt

$

$\displaystyle

\int t^2 \cdot e^{-2t} \, dt

$

now use parts ... in fact, tabular integration works well.