Evaluate the integral: integral (e^x dx)/(e^x + 2)

Evaluate the integral: $$ \int \frac{e^x \, dx}{e^x + 2}. $$

Explanation of Method

Since $\tfrac{d}{dx}(e^x+2)=e^x$ matches the numerator exactly, substituting $u=e^x+2$ reduces the integral to the pure logarithmic form $\int \frac{du}{u}$.

Substitute $u$ for the denominator.
Recognize that $du$ exactly matches the numerator.
Integrate using the standard logarithmic antiderivative.
Substitute back and add the constant of integration.

Step 1 · Substitution

We solve this integral using substitution. Let:

$$ u = e^x + 2. $$

Then, differentiate both sides with respect to $x$:

$$ \frac{du}{dx} = e^x \quad \Rightarrow \quad du = e^x\,dx. $$

Step 2 · Substitute and integrate

Now, substitute into the integral:

$$ \int \frac{e^x\,dx}{e^x+2} = \int \frac{du}{u}. $$

The integral of $\dfrac{1}{u}$ is a standard result:

$$ \int \frac{du}{u} = \ln|u|. $$

Step 3 · Substitute back

Substitute back $u = e^x+2$ to return to the variable $x$:

$$ \ln\left|e^x+2\right| + C. $$

Summary Table

Quantity	Value
Substitution	$u = e^x+2$
$du$	$e^x\,dx$ — matches the numerator exactly
Antiderivative	$\ln\|u\|$

Additional Notes

The integrand is defined for all real $x$, since $e^x+2 \geq 2 > 0$ always (so the absolute value is not strictly necessary here).
This is another instance of the "numerator equals derivative of denominator" pattern seen in Problem 67, which always signals a direct logarithmic antiderivative.
The result can be checked by differentiating $\ln\left(e^x+2\right)$ with the chain rule and confirming it returns $\dfrac{e^x}{e^x+2}$.

Result

The solution to the integral is:

$$ \ln\left|e^x+2\right| + C. $$