Answer:
[tex]I=\dfrac{1}{2}\dfrac{\ln \left|e^x+1\right|}{\ln \left|e^x-1\right|}+C[/tex]
Step-by-step explanation:
The given in integral problem is
[tex]\int \dfrac{e^x}{1-e^{2x}}dx[/tex]
[tex]\int \dfrac{e^x}{1-(e^{x})^2}dx[/tex]
Substitute [tex]e^x=t[/tex].
[tex]\dfrac{d}{dx}e^x=\dfrac{d}{dx}t[/tex]
[tex]e^x=\dfrac{dt}{dx}[/tex]
[tex]e^xdx=dt[/tex]
After substitution we get
[tex]\int \dfrac{t}{1-t^2}dt[/tex]
We know that
[tex]\int \dfrac{1}{a^2-x^2}dc=\dfrac{1}{2a}\dfrac{\ln \left|x+a\right|}{\ln \left|x-a\right|}+C[/tex]
Using this formula we get
[tex]\int \dfrac{1}{1^2-t^2}du=\dfrac{1}{2(1)}\dfrac{\ln \left|t+1\right|}{\ln \left|t-1\right|}+C[/tex]
Substitute [tex]t=e^x[/tex].
[tex]I=\dfrac{1}{2}\dfrac{\ln \left|e^x+1\right|}{\ln \left|e^x-1\right|}+C[/tex]
Therefore, the solution of given integral is [tex]I=\dfrac{1}{2}\dfrac{\ln \left|e^x+1\right|}{\ln \left|e^x-1\right|}+C[/tex].
