Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

I need to know if the improper integral from 1 to + infinity ( ln x / ( x + ln (x + 1) ) dx converges. Help?

OpenStudy (anonymous):

use limit x -> inf and rewrite the integral, solve using limits and then integrate the result

OpenStudy (anonymous):

\[\int\limits_{1}^{+\infty} \left( \frac{ \ln }{ x + \ln x } \right)\]

OpenStudy (anonymous):

Even if the boundaries are 1 and c, I still can't integrate.

OpenStudy (turingtest):

\[\int_1^\infty\frac{\ln xdx}{x+\ln(x+1)}\]?

zepdrix (zepdrix):

Hmm I can't figure this one out, are you suppose to use power series or something? :\

OpenStudy (turingtest):

you wrote two different things

OpenStudy (turingtest):

I was thinking perhaps we could use a comparison test if we are clever, but is the problem as you typed it in the post, or as you typed it below?

OpenStudy (anonymous):

Turing test, yes, what you wrote.

OpenStudy (anonymous):

use concept of improper integrals,

OpenStudy (anonymous):

As I typed it in the problem.

OpenStudy (anonymous):

\[\lim_{t \rightarrow \infty} \int\limits_{1}^{t} f(x)dx\]

OpenStudy (anonymous):

i have modified the formula, but put value of f(x)=function you typed in the question

OpenStudy (turingtest):

yeah, and how do you integrate that sucker?

OpenStudy (anonymous):

...

OpenStudy (anonymous):

multiple ways, try adding 1 and taking away one from numerator or double substitutiion

OpenStudy (anonymous):

@TuringTest , clearly you use the concept of integration to integrate it.

OpenStudy (anonymous):

double sub is a long method but will get you the answer, if you want to avoid the former method

OpenStudy (turingtest):

how about a demonstration? what substitution

OpenStudy (anonymous):

yeah show us your method @psi9epsilon

OpenStudy (turingtest):

I say we look for a comparison, but I can't think of one... I have to leave unfortunately

OpenStudy (anonymous):

try comparing to the integral of ln(x) /2x

OpenStudy (turingtest):

nice @Algebraic! that should do the trick :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!