Integral Exercise 33

We are tasked with evaluating the integral:

$$ \int \frac{e^{2x}}{3 - e^x} \ dx $$

We begin by rewriting the integrand using properties of exponents:

$$ \int \frac{e^x \cdot e^x}{3 - e^x} \ dx = \int \frac{e^x}{3 - e^x} \cdot e^x \ dx $$

Next, we apply the substitution method.

Let $ u = e^x $. Then the differential becomes:

$$ du = e^x \ dx $$

This allows us to replace $ e^x \ dx $ with $ du $:

$$ \int \frac{e^x}{3 - e^x} \cdot e^x \ dx = \int \frac{e^x}{3 - e^x} \ du $$

Substituting $ u = e^x $, the integral becomes:

$$ \int \frac{u}{3 - u} \ du $$

To simplify the integrand, we factor out a minus sign from the denominator:

$$ = - \int \frac{-u}{3 - u} \ du $$

Now we rewrite the numerator by adding and subtracting 3:

$$ = - \int \frac{-u + 3 - 3}{3 - u} \ du $$

This splits into two simpler terms:

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

$$ = - \left[ \int \frac{-3}{3 - u} \ du + \int 1 \ du \right] $$

The second integral is straightforward:

$$ \int 1 \ du = u $$

So we now have:

$$ - \left[ -3 \int \frac{1}{3 - u} \ du + u \right] + c $$

We evaluate the remaining integral by substituting $ t = 3 - u $, hence:

$$ dt = -du \quad \Rightarrow \quad du = -dt $$

Substituting, we get:

$$ - \left[ -3 \int \frac{1}{t} \cdot (-1) \ dt + u \right] + c $$

$$ = - \left[ 3 \int \frac{1}{t} \ dt + u \right] + c $$

The integral of $ 1/t $ is standard:

$$ \int \frac{1}{t} \ dt = \log |t| $$

Therefore:

$$ - \left[ 3 \log |t| + u \right] + c $$

Now substitute back $ t = 3 - u $:

$$ - \left[ 3 \log |3 - u| + u \right] + c $$

Which simplifies to:

$$ -3 \log |3 - u| - u + c $$

Finally, substituting $ u = e^x $ yields the solution:

$$ -3 \log |3 - e^x| - e^x + c $$

And that completes the evaluation of the integral.