# Integration by Parts

• Feb 22nd 2013, 03:31 PM
itshayley
Integration by Parts
Hi all, I'm a little new here, but I hope you don't mind me joining your little community as I delve into Calculus II. I'd really like to learn to enjoy this class and be proficient in it. I hope you all will support me in this.

Directions: In Exercises 11-38, find the integral. (Note: Solve by the simplest method--not all require integration by parts.)

Attachment 27195

I tried making
dv=ln2x
v=1/2x
u=x^-2
du=-2/x^3
When plugging it into integral-calculator and Wolfphram Alpha I get nothing like my answer which was:

1/2x^3 + -1/4x^3 + c
• Feb 22nd 2013, 04:01 PM
chiro
Re: Integration by Parts
Hey itshayley.

The derivative of ln(2x) is 1/x and it isn't the integral.

Try setting up dv = ln(2x) and u = x^(-2) [You have the v variables the wrong way around].
• Feb 23rd 2013, 04:49 AM
itshayley
Re: Integration by Parts
Thank you chiro. But my understanding was that v = the integral of dv not the derivative.
• Feb 23rd 2013, 02:58 PM
chiro
Re: Integration by Parts
The integral of ln(2x) is not 1/2x.
• Feb 23rd 2013, 11:14 PM
hollywood
Re: Integration by Parts
For $\int\frac{\ln{2x}}{x^2}\,dx$, you should let $u=\ln{2x}$ and $dv=\frac{1}{x^2}\,dx$. Then $du=\frac{1}{x}\,dx$ and $v=-\frac{1}{x}$. So:

$\int\frac{\ln{2x}}{x^2}\,dx =$

$-\frac{1}{x}\ln{2x}+\int\frac{1}{x}\frac{1}{x}\,dx =$

$-\frac{\ln{2x}}{x}-\frac{1}{x}$

There is a way to remember what to put in u and what to put in dv - it's "LIATE" - you choose for u whatever comes first in the list Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. It's not foolproof, but it's a pretty good rule of thumb.

- Hollywood
• Feb 24th 2013, 12:42 AM
ibdutt
Re: Integration by Parts
• Feb 24th 2013, 08:38 AM
itshayley
Re: Integration by Parts
Wow, that was so much simpler than I was making it. Thanks Hollywood. I will keep that tip in mind.

Thank you ibdutt for your input as well.

I see now chiro, sorry for the miscommunication.