Evaluating Integral $$\int \frac{e^{2x}}{u} du$$

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Discussion Overview

The discussion revolves around evaluating the integral $$\int \frac{e^{2x}}{e^{2x}-2}dx$$ using a substitution method. Participants explore the implications of their substitutions and the resulting expressions, focusing on the correctness of the differential and the integration process.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes a substitution with $$u=e^{2x}-2$$ and calculates the differential as $$du=e^{2x}$$.
  • Another participant corrects the differential, stating it should be $$du=2e^{2x}\,dx$$.
  • A third participant suggests that the integral simplifies to $$\int \frac{1}{u}\ du\ dx = \ln\left({e^{2x}-2}\right)/2$$.
  • A later reply reiterates the integral result, emphasizing the correct form as $$\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln\left|e^{2x}-2\right|+C$$.

Areas of Agreement / Disagreement

Participants express differing views on the calculation of the differential and the integration steps, indicating that there is no consensus on the correct approach or final expression.

Contextual Notes

The discussion highlights potential limitations in the initial substitution and the calculation of differentials, which may affect the integration process. The assumptions regarding the substitution and the handling of logarithmic expressions remain unresolved.

karush
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$$\int_{}^{} \frac{e^{2x}}{e^{2x}-2}dx. \\u=e^{2x}-2\\du=e^{2x}$$
Now what?
 
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Your $u$-substitution is good, but you have calculated the differential incorrectly. Given:

$$u=e^{2x}-2$$

Then:

$$du=2e^{2x}\,dx$$
 
$\int_{}^{}\frac{1}{u}\ du\ dx = \ln\left({e^{2x}-2}\right)/2$
 
karush said:
$\int_{}^{}\frac{1}{u}\ du\ dx = \ln\left({e^{2x}-2}\right)/2$

You actually want:

$$\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|+C=\frac{1}{2}\ln\left|e^{2x}-2\right|+C$$
 

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