# Integration By Parts-Is my answer correct?

∫9(ln9x)2dx

## 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

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

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?

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

jgens
Gold Member
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
Gold Member
9x(ln9x)2-18xln9x+18x+C

So, is this correct or where is my error?
This is the answer that I got.

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

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

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
Gold Member
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.

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.

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.

Not true. The derivative is in fact 1/x if you apply the chain rule correctly.
Yeah, um, wow. [sackcloth and ashes].

HallsofIvy
Science Advisor
Homework Helper
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.

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