Integration By Parts-Is my answer correct?

  • Thread starter neshepard
  • Start date
  • #1
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Homework Statement



∫9(ln9x)2dx

Homework Equations





The Attempt at a Solution


u=(ln9x)2 dv=9dx
du=2(ln9x)*1/xdx v=9x

After all my work I get:
9x(ln9x)2-18(ln9x)+18x

Webassign tells me I'm wrong, but I have worked it 3 times to the same answer.
Please help me.
Neil
 

Answers and Replies

  • #2
351
1
Your substitution is all right. Check du, however. You have to use the chain rule twice.
 
  • #3
67
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I did a 2nd substitution, but I was trying to save writing out the lines. Here is my work:
9x(ln9x)2-18∫xln9x*1/xdx
9x(ln9x)2-18∫ln9x dx (here the x * 1/x cancelled)

u=ln9x dv=dx
du=1/xdx v=x

9x(ln9x)2-18[xln9x-∫dx]
9x(ln9x)2-18[xln9x-x]

=9x(ln9x)2-18xln9x+18x+C

So, is this correct or where is my error?
 
  • #4
351
1
The derivative of ln 9x is not 1/x, it's 9/x by the chain rule.
 
  • #5
jgens
Gold Member
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The derivative of ln 9x is not 1/x, it's 9/x by the chain rule.
Not true. The derivative is in fact 1/x if you apply the chain rule correctly.
 
  • #6
jgens
Gold Member
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9x(ln9x)2-18xln9x+18x+C

So, is this correct or where is my error?
This is the answer that I got.
 
  • #7
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Cool, then it's the fact that webassign hates a Mac. Thanks
Neil
 
  • #8
put u=9x
du/9=dx
so integral (lnu)2 du
put v=ln(u)
dv=du/u
since u=ev
so
evdv=du

integral v2 evdv
v2ev-2vev+2ev+c
so 9x(ln(9x))2-18xln9x+18x+c
 
  • #9
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After all my work I get:
9x(ln9x)2-18(ln9x)+18x

Webassign tells me I'm wrong, but I have worked it 3 times to the same answer.
Please help me.
Neil
The bolded part should be 18x. Maybe you put in 18 instead of 18x so you lost points for it.
 
  • #10
jgens
Gold Member
1,581
50
put u=9x
du/9=dx
so integral (lnu)2 du
put v=ln(u)
dv=du/u
since u=ev
so
evdv=du

integral v2 evdv
v2ev-2vev+2ev+c
so 9x(ln(9x))2-18xln9x+18x+c
If you choose u=log(9x) from the start, you can reduce the number of computations that you need to reach exp(u)u2.
 
  • #11
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LEPTOS, in my work on paper I have 18x and I put that in webassign. I really!!!!can't type. My math is better than my typing if that says anything.
 
  • #12
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The very first set of u and du are not actually substitutions (or at least as I was told by my professor). Whom ever created our textbook could find a better letter combo in the entire 26 letters of the alphabet. Go figure. The use of u and du in integration by parts has caused me pain to no end because I want to pull the 2 in the du and get 1/2du=(ln9x)*1/xdx...but apparently that is wrong.
 
  • #13
351
1
Not true. The derivative is in fact 1/x if you apply the chain rule correctly.
Yeah, um, wow. [sackcloth and ashes].
 
  • #14
HallsofIvy
Science Advisor
Homework Helper
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956
You can find the derivative of f(x)= ln(9x) in either of two ways:

chain rule: let u= 9x. Then du/dx= 9 and f(u)= ln(u) so df/du= 1/u= 1/(9x). Now df/dx= (df/du)(du/dx)= (9)(1/(9x))= 1/x.

The easy way: f(x)= ln(9x)= ln(x)+ ln(9). Since ln(9) is a constant, df/dx= d(ln(x))/dx+ 0= 1/x.
 
  • #15
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Just because I can say this, and laugh about it...HallsofIvy....the easy way....calculator. j/k
 

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