Solve Tricky Integration: ##\frac{20\ln{(t)}}{t}##

[squenshl]

Homework Statement

How do I solve $\frac{20\ln{(t)}}{t}$?

Homework Equations

The Attempt at a Solution

Is it easier to calculate this without integrating by parts?
I'm not sure where to start.

 

[andrewkirk]

Is it easier to calculate this without integrating by parts?

Solving by integration by parts is easy - very few lines. I can't think of an easier way.

 

[Tallus Bryne]

There is an easier way. Try a simple "u-substitution".

 

[fresh_42]

You have $f\cdot f'$ which is half of $(f^2)'$. Done.

 

[Tallus Bryne]

You have $f\cdot f'$ which is half of $(f^2)'$. Done.

I like the simplicity, but shouldn't it be just half of $(f^2)$ ?

edited: forgot the "half of"

 

[fresh_42]

I like the simplicity, but shouldn't it be just half of $(f^2)$ ?

edited: forgot the "half of"

$f^2$ is the solution after integration, but I said $f \cdot f' = \frac{1}{2} \cdot (f^2)'$ for the integrand. Of course this all is a bit sloppy: no integration boundaries or the constant, no mentioning of $\ln |x|$ in the OP and no $dt 's$.

 

[Tallus Bryne]

$f^2$ is the solution after integration, but I said $f \cdot f' = \frac{1}{2} \cdot (f^2)'$ for the integrand. Of course this all is a bit sloppy: no integration boundaries or the constant, no mentioning of $\ln |x|$ in the OP and no $dt 's$.

Agreed. Thanks for clearing that up.

 

[member 587159]

Substitute $u = \ln t, du = \frac {1}{t}dt$

 

[squenshl]

Substitute $u = \ln t, du = \frac {1}{t}dt$

Thanks everyone. The solution is $10\ln{(t)}^2$.

 

[SammyS]

Thanks everyone. The solution is $10\ln{(t)}^2$.

You might like to remove an ambiguity.

Is that the natural log of the square of t, or is it the square of the natural log of t ?

Also, I suspect that you need to include a constant of integration .

 

[MidgetDwarf]

Typically for integrals involving ln in a typical calculus course...

Always try u sub. If that does not work, then integration by parts. Keep this in mind.