Funny question we came across in class... (1 Viewer)

[CJBrownie]

Maybe someone can help me work this out....

We had the following question:

Find the integral of 1/xlnx

Using the substitution u = lnx the question works out quite nicely.

However, inadvertanely missing this, I tried to do this question using integration by parts, and recieved a rather startling result.

If you have the time, try doing this question by parts as follows, and in about 20 seconds you will realise the problem I had.

Let u = 1/lnx

Let dV/dx = 1/x

 

[FinalFantasy]

u probably got into some problem cuz it's inappropriate to use u=1\lnx and dv\dx=1\x for that integral..

 

[CJBrownie]

Did you do the problem though and see what the error is?

 

[FinalFantasy]

Find the integral of 1/xlnx
let I=int. 1\xlnx

let u=1\lnx and dV\dX=1\x
du\dx=-(1\x)\(lnx)² v=lnx

I=1+int. lnx(1\x(lnx)²)
=1+int. 1\xlnx=1+I
I=1+I
0=1 lol

is dat da "problem"?

 

[SeDaTeD]

Yay 0=1
Just chuck a plus C on one side and ull be fine hehe

 

[CJBrownie]

Hehe that's the problem.

The original question had limits to substitute into the answer, so the problem being a constant wasn't considered.

 

[CJBrownie]

I worked it out!

Sadly enough.... 0 does not equal 1.

let I=int. 1\xlnx

let u=1\lnx and dV\dX=1\x
du\dx=-(1\x)\(lnx)² v=lnx

I=1+int. lnx(1\x(lnx)²) ----- this is the problem line
=1+int. 1\xlnx=1+I
I=1+I
0=1


The question had limits of between e and e^2.

So that line was

I = [1] + int. [1\xlnx]

(where [ ] means between e and e^2)

Most people who have done the question have assumed that [1] = 1, because there are no variables to substitute the limits into.

but [1] = 1 - 1 = 0

so the question from that point becomes:

I = [1] + int. [1\xlnx]
I =0 + I
I=I

Which is obviously true.