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

[karush]

$$\int_{}^{} \frac{e^{2x}}{e^{2x}-2}dx. \\u=e^{2x}-2\\du=e^{2x}$$
Now what?

 

[MarkFL]

Your $u$-substitution is good, but you have calculated the differential incorrectly. Given:

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

Then:

$$du=2e^{2x}\,dx$$

 

[karush]

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

 

[MarkFL]

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

You actually want:

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