Transcendental Functions Containing Natural Logs

[WhiskeyHammer]

Homework Statement

Evaluate the integrals

a) y=

b) y=

Homework Equations

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 haven't 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 I am at a loss of how to fix it.

 

[LCKurtz]

You want the anti-derivative (definite integral), not the derivative, and there is no $y$ in either problem. In both problems try the substitution $u=\ln x$.

 

[Mark44]

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.

 

[WhiskeyHammer]

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 don't 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.

 

[Mark44]

I see the intent in setting up the integral as $u*du$ and anti-deriving to $2u$ but I don't 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.

 

[HallsofIvy]

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 $(1/2)u^2+ C$

I don't 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
$$\int \frac{dx}{x ln(x)}$$
(the second "dx" must be a typo.)

If you let u= ln(x) then $du= \frac{1}{x}dx= \frac{dx}{x}$ so the integral becomes
$$\int\frac{dx}{x ln(x)}= \int \frac{1}{ln(x)}\frac{dx}{x}= \int \frac{1}{u}du$$

"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.

 

[WhiskeyHammer]

Thanks guys, despite the errors on my part, you suggestions and clarifications were awesome and I got it worked out