Integral Exercise 35

We are asked to evaluate the following integral:

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

First, observe that $ e^{x+1} = e^x \cdot e $, which allows us to factor out the constant $ e $:

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

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

To proceed, we'll apply the substitution method.

Let’s differentiate the expression $ 3 + e^x $:

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

Solving for $ dx $, we get:

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

Substituting this into the integral:

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

The $ e^x $ terms cancel out, leaving:

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

We now set $ t = 3 + e^x $, so the integral becomes:

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

This is a standard integral, as $ \int \frac{1}{t} \, dt = \log |t| + c $:

$$ e \cdot \log |t| + c $$

Substituting back for $ t = 3 + e^x $, we obtain:

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

Therefore, the final result is:

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

And so on.