Evaluate the integral: I = integral (e^x)/(sqrt(e^x+1)) dx
Evaluate the integral: $$ I = \int \frac{e^x}{\sqrt{e^x+1}} \, dx. $$
Explanation of Method
Since $\tfrac{d}{dx}(e^x+1)=e^x$ matches the numerator exactly, substituting $u=e^x+1$ turns the square root into $\sqrt u$ and reduces the integral to a routine power-rule antiderivative.
- Substitute $u$ for the expression under the root.
- Recognize that $du$ matches the numerator exactly.
- Integrate the resulting fractional power of $u$.
- Substitute back and add the constant of integration.
Step 1 · Substitution
We use the substitution $u = e^x+1$, so that:
$$ du = e^x\,dx. $$
Step 2 · Rewrite the integral
Rewriting the integral in terms of $u$, we get:
$$ I = \int \frac{du}{\sqrt{u}}. $$
This simplifies to:
$$ I = \int u^{-1/2}\,du. $$
Step 3 · Integrate
Using the standard integral formula:
$$ \int u^n\,du = \frac{u^{n+1}}{n+1}, \quad \text{for } n\neq -1, $$
with $n=-\tfrac12$, we obtain:
$$ I = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C. $$
Step 4 · Substitute back
Substituting back $u = e^x+1$, we get:
$$ I = 2\sqrt{e^x+1} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = e^x+1$ |
| $du$ | $e^x\,dx$ — matches the numerator exactly |
| Antiderivative | $2\sqrt{u}$ |
Additional Notes
- The integrand is defined for all real $x$, since $e^x+1 > 0$ always.
- This is one of the simplest possible substitutions: the numerator exactly equals the derivative of the expression under the radical, so no leftover constant needs to be tracked.
- The result can be checked by differentiating $2\sqrt{e^x+1}$ with the chain rule and confirming it returns $\dfrac{e^x}{\sqrt{e^x+1}}$.
Result
The solution to the integral is: