
1 Attachment(s)
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 1138, find the integral. (Note: Solve by the simplest methodnot 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 integralcalculator and Wolfphram Alpha I get nothing like my answer which was:
1/2x^3 + 1/4x^3 + c

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].

Re: Integration by Parts
Thank you chiro. But my understanding was that v = the integral of dv not the derivative.

Re: Integration by Parts
The integral of ln(2x) is not 1/2x.

Re: Integration by Parts
For $\displaystyle \int\frac{\ln{2x}}{x^2}\,dx$, you should let $\displaystyle u=\ln{2x}$ and $\displaystyle dv=\frac{1}{x^2}\,dx$. Then $\displaystyle du=\frac{1}{x}\,dx$ and $\displaystyle v=\frac{1}{x}$. So:
$\displaystyle \int\frac{\ln{2x}}{x^2}\,dx =$
$\displaystyle \frac{1}{x}\ln{2x}+\int\frac{1}{x}\frac{1}{x}\,dx = $
$\displaystyle \frac{\ln{2x}}{x}\frac{1}{x}$
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

1 Attachment(s)
Re: Integration by Parts

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.