How to Solve the Integral of 1/(x*ln(x)^2)?

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Homework Help Overview

The discussion revolves around solving the integral of the function 1/(x*ln(x)^2), which falls under the subject area of calculus, specifically integration techniques.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts a u-substitution with u = ln(x) and expresses confusion regarding the correctness of their antiderivative. Some participants suggest alternative methods, such as integration by parts, while others question the validity of the original poster's steps and encourage them to share their full solution for further assistance.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's approach and suggesting re-evaluation of the substitution process. There is no explicit consensus on the correct method, but several participants are engaged in clarifying the reasoning behind the steps taken.

Contextual Notes

Participants note that the original poster has not yet covered certain integration techniques, such as integration by parts, which may limit their approach to the problem.

derekjn
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Homework Statement



\int\frac{1}{xln(x)^2}

Homework Equations



\frac{d}{dx}ln(x) = \frac{1}{x}

The Attempt at a Solution



I made a u substitution, letting u = ln(x); du = \frac{1}{x}dx. The antiderivative I then get is ln(u^{2}) + C but obviously this is not the correct answer. Any help anyone could provide would be greatly appreciated.

Derek
 
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Have you tried integrating by parts?
 
Thanks for your reply. Unfortunately we haven't covered integration by parts yet; we have only made it as far as logarithmic differentiation so the techniques that I know are limited. derek
 
the procedure you are doing seems correct but your final answer is not correct. You are therefore making an error in the process. Please post your full solution and perhaps someone will point out the error. Or if this helps the answer should be

-1/lnx + c
 
Integration by parts is not necessary at all. Just rework your problem with the substitution you chose, and you'll see the error you are making.
 
Thanks for your replies guys. I'm still not seeing it. After the substitution I get \int\frac{1}{u^{2}}du = ln(u^{2}) + C = 2ln(u) + C. I'm stumped at this point.
 
\int x^{-2} dx = -1/x.
 
Beautiful. I got so caught up in \frac{d}{dx}ln(x) = \frac{1}{x} that the square didn't even register. Thanks everyone for your help. Derek
 
answer comes out to be 1/2(ln(ln(x^2))+c
 

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