Integration By Parts-Is my answer correct?

[neshepard]

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

 

[hgfalling]

Your substitution is all right. Check du, however. You have to use the chain rule twice.

 

[neshepard]

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?

 

[hgfalling]

The derivative of ln 9x is not 1/x, it's 9/x by the chain rule.

 

[jgens]

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.

 

[jgens]

9x(ln9x)²-18xln9x+18x+C

So, is this correct or where is my error?

This is the answer that I got.

 

[neshepard]

Cool, then it's the fact that webassign hates a Mac. Thanks
Neil

 

[jamalahmed68]

put u=9x
du/9=dx
so integral (lnu)² du
put v=ln(u)
dv=du/u
since u=e^v
so
e^vdv=du

integral v² e^vdv
v²e^v-2ve^v+2e^v+c
so 9x(ln(9x))²-18xln9x+18x+c

 

[Leptos]

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.

 

[jgens]

put u=9x
du/9=dx
so integral (lnu)² du
put v=ln(u)
dv=du/u
since u=e^v
so
e^vdv=du

integral v² e^vdv
v²e^v-2ve^v+2e^v+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.

 

[neshepard]

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.

 

[neshepard]

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.

 

[hgfalling]

Not true. The derivative is in fact 1/x if you apply the chain rule correctly.

Yeah, um, wow. [sackcloth and ashes].

 

[HallsofIvy]

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.

 

[neshepard]

Just because I can say this, and laugh about it...HallsofIvy...the easy way...calculator. j/k