Evaluate the integral: I = integral (e^x+1)/(e^x+x) dx

Evaluate the integral: $$ I = \int \frac{e^x+1}{e^x+x} \, dx. $$

Explanation of Method

Even though the denominator mixes an exponential and a polynomial term, its derivative $\tfrac{d}{dx}(e^x+x)=e^x+1$ matches the numerator exactly — so the same logarithmic substitution technique applies regardless of how unusual the expression looks.

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 use the substitution $u = e^x + x$, so that:

$$ du = (e^x+1)\,dx. $$

Step 2 · Rewrite and integrate

Rewriting the integral in terms of $u$, we get:

$$ I = \int \frac{du}{u}. $$

This is a standard integral:

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

Step 3 · Substitute back

Substituting back $u = e^x+x$, we obtain:

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

Summary Table

Quantity	Value
Substitution	$u = e^x+x$
$du$	$(e^x+1)\,dx$ — matches the numerator exactly
Antiderivative	$\ln\|u\|$

Additional Notes

The domain of $I$ requires $e^x+x \neq 0$, which holds for all $x$ except at the single point where $e^x=-x$ (a solution near $x\approx-0.567$, the omega constant).
The fact that the denominator mixes an exponential and a linear term does not complicate the method at all — the "numerator equals denominator's derivative" test works the same way regardless of the function types involved.
The result can be checked by differentiating $\ln\left|e^x+x\right|$ with the chain rule and confirming it returns $\dfrac{e^x+1}{e^x+x}$.

Result

The solution to the integral is:

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