Integrate $\ln(t^2)/t$ with Substitution

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Discussion Overview

The discussion revolves around the integration of the function $\frac{\ln(t^2)}{t}$ with respect to $t$. Participants explore different approaches to the integral, including substitution methods and simplifications of logarithmic expressions.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the integral $\int\frac{\ln\left({t^2}\right)}{t}dt$ and attempts a substitution with $u=\ln(t)$, leading to $\frac{1}{2}\int u \, du$.
  • Another participant suggests a simplification, noting that $\frac{\ln(t^2)}{t}$ can be rewritten as $2 \frac{\ln(t)}{t}$, affirming the correctness of the evaluation up to that point.
  • A third participant points out that the logarithmic identity $\ln(x^2) = 2\ln(x)$ was misapplied, indicating that the factor of 2 was incorrectly placed in the denominator.
  • Subsequent replies confirm the adjustment, leading to the integral $2\int u \, du$ and the evaluation resulting in $(\ln(t))^2 + C$.
  • Participants agree that the factor of 2 cancels out in the final expression.

Areas of Agreement / Disagreement

Participants generally agree on the steps taken to simplify and evaluate the integral, with some minor corrections regarding the placement of factors. However, the discussion does not resolve any broader implications or alternative methods for integration.

Contextual Notes

The discussion does not address potential limitations or assumptions regarding the integration process or the domain of the logarithmic function.

karush
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$\int\frac{\ln\left({t^2}\right)}{t}dt$

$\frac{1}{2}\int\ln\left({t}\right)\frac{1}{t}dt$

$u=\ln\left({t}\right)\ du=\frac{1}{t}dt$

$\frac{1}{2}\int u\ du$I continued but didn't get the answer
 
Last edited:
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Hint: $\frac{\ln(t^2)}{t}= 2 \frac{\ln(t)}{t}$.
The rest of your evaluation is correct.
 
$\ln(x^2) = 2\ln(x)$. You put the 2 in the denominator.
 
So its.
$2\int\ u \ du$

$2 \left[\frac{u^2}{2}\right]+C$
 
Yep, and the 2's would cancel.
 
$(\ln\left({t}\right))^2+C$
 

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