Evaluate the integral: integral (e^t)/((e^t-3)(e^t+3)) dt

Let $u = e^t$.
Then, differentiate $u$ with respect to $t$:

$$du = e^t \, dt$$

Substitute $u$ and $du$ into the integral:

$$\int \frac{e^t}{(e^t-3)(e^t+3)} \, dt = \int \frac{1}{(u-3)(u+3)} \, du$$

We want to find constants $A$ and $B$ such that:

$$\frac{1}{(u-3)(u+3)} = \frac{A}{u-3} + \frac{B}{u+3}$$

Multiply both sides by $(u-3)(u+3)$:

$$1 = A(u+3) + B(u-3)$$

To find $A$, set $u=3$:

$$1 = A(3+3) + B(3-3) \implies 1 = 6A \implies A = \frac{1}{6}$$

To find $B$, set $u=-3$:

$$1 = A(-3+3) + B(-3-3) \implies 1 = -6B \implies B = -\frac{1}{6}$$

So, the partial fraction decomposition is:

$$\frac{1}{(u-3)(u+3)} = \frac{1/6}{u-3} - \frac{1/6}{u+3}$$

$$\int \left( \frac{1/6}{u-3} - \frac{1/6}{u+3} \right) \, du = \frac{1}{6} \int \frac{1}{u-3} \, du - \frac{1}{6} \int \frac{1}{u+3} \, du$$

The integral of $\frac{1}{x}$ is $\ln|x|$.

$$= \frac{1}{6} \ln|u-3| - \frac{1}{6} \ln|u+3| + C$$

Using the logarithm property $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:

$$= \frac{1}{6} \left( \ln|u-3| - \ln|u+3| \right) + C$$ $$= \frac{1}{6} \ln\left|\frac{u-3}{u+3}\right| + C$$

Now, substitute back $u=e^t$:

$$= \frac{1}{6} \ln\left|\frac{e^t-3}{e^t+3}\right| + C$$

The final answer is:

$$\boxed{\frac{1}{6} \ln\left|\frac{e^t-3}{e^t+3}\right| + C}$$

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