Solved Exercise: Integral Example 4

Let's solve this indefinite integral:

$$ \int \frac{e^{x+1}}{3+e^x} \, dx $$

Step-by-step solution:

First, we rewrite the integral in an equivalent form, noting that \( e^{x+1} = e^x \cdot e \):

$$ \int \frac{e^x \cdot e}{3+e^x} \, dx $$

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

To evaluate this integral, we'll use the substitution method.

Let’s consider the differential of \( 3 + e^x \):

$$ d(3+e^x) = e^x \, dx $$

which gives:

$$ dx = \frac{d(3+e^x)}{e^x} $$

Now, we substitute \( dx = \frac{d(3+e^x)}{e^x} \) into the integral:

$$ e \cdot \int \frac{e^x}{3+e^x} \, dx $$

$$ = e \cdot \int \frac{e^x}{3+e^x} \cdot \frac{d(3+e^x)}{e^x} $$

$$ = e \cdot \int \frac{d(3+e^x)}{3+e^x} $$

$$ = e \cdot \int \frac{1}{3+e^x} \, d(3+e^x) $$

Now, we introduce the substitution \( t = 3 + e^x \):

$$ = e \cdot \int \frac{1}{t} \, dt $$

This simplifies to the integral \( \int \frac{1}{t} \, dt = \log |t| + C \):

$$ = e \cdot \log |t| + C $$

Recalling that \( t = 3 + e^x \):

$$ e \cdot \log |t| + C = e \cdot \log |3 + e^x| + C $$

Therefore, the solution to the integral is:

$$ \int \frac{e^{x+1}}{3+e^x} \, dx = e \cdot \log |3 + e^x| + C $$