So I have the following problem:

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx\)

I realize this can be computed a simple U-substitution (followed by rationalizing the function), however I wanted to get in some practice with trig substitutions.

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^3} dx =\int \frac{4e^x}{(1-(e^x)^2)^3} dx \)

Now I set the following:

\(\displaystyle e^x = sin(t) \rightarrow x = ln(sin(t)) \rightarrow dx = \frac{1}{sin(t)} * cos(t) dt\)

Substituting:

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4\int \frac{sin(t)}{(1-sin(t)^2)^3} * \frac{1}{sin(t)} * cos *dt = 4\int \frac{1}{cos(t)^5}\)

\(\displaystyle 4\int sec(t)^5 dt = 4\int sec(t)^3 sec(t)^2 = sec(t)^3 tan(t) - \int 3 sec(t)^2 tan(t) dt\)

So we get:

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4(sec(t)^3 tan - \frac{3}{2}tan(t)^2 + C\)

To express it in terms of X:

\(\displaystyle cos(t) = \sqrt{1-(e^x)^2}\rightarrow sec(t) = \frac{1}{\sqrt{1-(e^x)^2}} \rightarrow tan(t) = \frac{e^x}{\sqrt{1-(e^x)^2}} \)

Finally substituting back, the final answer becomes:

\(\displaystyle 4((\frac{1}{\sqrt{1-(e^x)^2}})^3 \frac{e^x}{\sqrt{1-(e^x)^2}} - \frac{3}{2}(\frac{e^x}{\sqrt{1-(e^x)^2}})^2) + C\)

My problem is that the domain of this function is only x < 0 (for real numbers). What should I make of that?

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx\)

I realize this can be computed a simple U-substitution (followed by rationalizing the function), however I wanted to get in some practice with trig substitutions.

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^3} dx =\int \frac{4e^x}{(1-(e^x)^2)^3} dx \)

Now I set the following:

\(\displaystyle e^x = sin(t) \rightarrow x = ln(sin(t)) \rightarrow dx = \frac{1}{sin(t)} * cos(t) dt\)

Substituting:

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4\int \frac{sin(t)}{(1-sin(t)^2)^3} * \frac{1}{sin(t)} * cos *dt = 4\int \frac{1}{cos(t)^5}\)

\(\displaystyle 4\int sec(t)^5 dt = 4\int sec(t)^3 sec(t)^2 = sec(t)^3 tan(t) - \int 3 sec(t)^2 tan(t) dt\)

So we get:

\(\displaystyle \int \frac{4e^x}{(1-e^{2x})^2} dx = 4(sec(t)^3 tan - \frac{3}{2}tan(t)^2 + C\)

To express it in terms of X:

\(\displaystyle cos(t) = \sqrt{1-(e^x)^2}\rightarrow sec(t) = \frac{1}{\sqrt{1-(e^x)^2}} \rightarrow tan(t) = \frac{e^x}{\sqrt{1-(e^x)^2}} \)

Finally substituting back, the final answer becomes:

\(\displaystyle 4((\frac{1}{\sqrt{1-(e^x)^2}})^3 \frac{e^x}{\sqrt{1-(e^x)^2}} - \frac{3}{2}(\frac{e^x}{\sqrt{1-(e^x)^2}})^2) + C\)

My problem is that the domain of this function is only x < 0 (for real numbers). What should I make of that?

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