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Transcendental Functions Containing Natural Logs

  • #1

Homework Statement


Evaluate the integrals

a) y=
eqn40561.png

b) y=
eqn4056.png


Homework Equations



Untitled-1.png


The Attempt at a Solution



I've tried using the Reciprocal Rule and substitution of each ln function with u to get a workable result, but havent come anywhere near a recognizable solution. The answers stated in the book are as follows:
a) 1/ln4
b) (ln2)^2

[STRIKE]This tells me that, somehow a must end up looking like this:
ln 1 - ln 4
and b must look like:
ln2 * ln2[/STRIKE]

These would seem counter intuitive to my understanding of how these kinds of problems are solved and Im at a loss of how to fix it.
 
Last edited:

Answers and Replies

  • #2
LCKurtz
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You want the anti-derivative (definite integral), not the derivative, and there is no ##y## in either problem. In both problems try the substitution ##u=\ln x##.
 
  • #3
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The answers stated in the book are as follows:
a) 1/ln4


This tells me that, somehow a must end up looking like this:
ln 1 - ln 4
1/ln(4) ≠ ln(1) - ln(4), so it would be good for you to review the properties of logs.

The expression on the right can be simplified like so:

ln(1) - ln(4) = 0 - ln(4) = ln(1/4)

Note the ln(1/4) is a negative number, while 1/ln(4) is positive, so they couldn't possibly be equal.
 
  • #4
You want the anti-derivative (definite integral), not the derivative, and there is no ##y## in either problem. In both problems try the substitution ##u=\ln x##.
I see the intent in setting up the integral as ##u*du## and anti-deriving to ##2u## but I dont see how ##du## could possibly equal ##1/x*dx## the derivative of ##\ln x## is ##(1/x)(d/dx*x)## which leaves us with ##u*du*dx## right?

I also updated the errors in my original post. Thanks for catching them, I am definitely on the tail end of burned out.
 
  • #5
33,075
4,779
I see the intent in setting up the integral as ##u*du## and anti-deriving to ##2u## but I dont see how ##du## could possibly equal ##1/x*dx## the derivative of ##\ln x## is ##(1/x)(d/dx*x)## which leaves us with ##u*du*dx## right?
No. If u = lnx, then du = d(lnx) = d/dx(lnx)*dx = 1/x * dx
I also updated the errors in my original post. Thanks for catching them, I am definitely on the tail end of burned out.
 
  • #6
HallsofIvy
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I see the intent in setting up the integral as ##u*du## and anti-deriving to ##2u## but
Ouch! The anti-derivative of u is [itex](1/2)u^2+ C[/itex]

I dont see how ##du## could possibly equal ##1/x*dx## the derivative of ##\ln x## is ##(1/x)(d/dx*x)## which leaves us with ##u*du*dx## right?

I also updated the errors in my original post. Thanks for catching them, I am definitely on the tail end of burned out.
Your original integral was
[tex]\int \frac{dx}{x ln(x)}[/tex]
(the second "dx" must be a typo.)

If you let u= ln(x) then [itex]du= \frac{1}{x}dx= \frac{dx}{x}[/itex] so the integral becomes
[tex]\int\frac{dx}{x ln(x)}= \int \frac{1}{ln(x)}\frac{dx}{x}= \int \frac{1}{u}du[/tex]

"the derivative of ##\ln x## is ##(1/x)(d/dx*x)## "
I can't make sense of that! The derivative of ln(x) is 1/x. Perhaps you mean (1/x) times the derivative of x with respect to x. That last would be 1 so that is the same as 1/x. Having written du/dx= 1/x, in differential terms, du= (1/x)dx.
 
  • #7
Thanks guys, despite the errors on my part, you suggestions and clarifications were awesome and I got it worked out :biggrin:
 

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