Evaluate the integral: I = integral dx/(sqrt(1+e^2x))
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt{1+e^{2x}}}. $$
Explanation of Method
Multiplying the integrand by $e^{-x}/e^{-x}$ moves the exponential growth into a form where the substitution $w=e^{-x}$ produces the clean standard integral $\int \frac{dw}{\sqrt{1+w^2}}$, which has a known logarithmic (inverse hyperbolic sine) antiderivative.
- Multiply numerator and denominator by $e^{-x}$.
- Substitute $w=e^{-x}$.
- Apply the standard $\int \frac{dw}{\sqrt{1+w^2}}$ antiderivative.
- Substitute back and simplify.
Step 1 · Rewrite the integrand
Multiply the numerator and denominator by $e^{-x}$:
$$ I = \int \frac{dx}{\sqrt{1+e^{2x}}} = \int \frac{e^{-x}\,dx}{e^{-x}\sqrt{1+e^{2x}}}. $$
Since $e^{-x}\sqrt{1+e^{2x}} = \sqrt{e^{-2x}(1+e^{2x})} = \sqrt{1+e^{-2x}}$, this becomes:
$$ I = \int \frac{e^{-x}\,dx}{\sqrt{1+e^{-2x}}}. $$
Step 2 · Substitute
Let $w = e^{-x}$. Then $dw = -e^{-x}\,dx$, so $e^{-x}\,dx = -dw$. The integral becomes:
$$ I = \int \frac{-dw}{\sqrt{1+w^2}} = -\int \frac{dw}{\sqrt{1+w^2}}. $$
Step 3 · Apply the standard result
The integral $\displaystyle \int \frac{dw}{\sqrt{1+w^2}}$ is a standard result:
$$ \int \frac{dw}{\sqrt{1+w^2}} = \ln\left(w+\sqrt{1+w^2}\right). $$
Thus:
$$ I = -\ln\left(w+\sqrt{1+w^2}\right) + C. $$
Step 4 · Substitute back
Substituting back $w = e^{-x}$, we get:
$$ I = -\ln\left(e^{-x}+\sqrt{1+e^{-2x}}\right) + C. $$
This can be simplified: factoring $e^{-x}$ out of the argument, $e^{-x}+\sqrt{1+e^{-2x}} = e^{-x}\left(1+\sqrt{1+e^{2x}}\right)$, so:
$$ I = -\ln\left(e^{-x}\right) - \ln\left(1+\sqrt{1+e^{2x}}\right) + C = x - \ln\left(1+\sqrt{1+e^{2x}}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $w = e^{-x}$ |
| Reduced integral | $-\int dw/\sqrt{1+w^2}$ |
| Antiderivative | $x - \ln\left(1+\sqrt{1+e^{2x}}\right)$ |
Additional Notes
- The integrand is defined for all real $x$, since $1+e^{2x} > 0$ always.
- A naive substitution $t=e^x$ followed by $u=\ln\left(t+\sqrt{1+t^2}\right)$ can look tempting, but $du = \dfrac{dt}{\sqrt{1+t^2}}$ leaves an extra factor of $\dfrac1t$ unaccounted for in the integral $\int \dfrac{dt}{t\sqrt{1+t^2}}$ — treating that integral as simply $\int du$ is a common error, since the $\dfrac1t$ factor does not cancel. The $w=e^{-x}$ route avoids this pitfall entirely.
- The result can be checked by differentiating $x-\ln\left(1+\sqrt{1+e^{2x}}\right)$ with the chain rule and confirming it returns $\dfrac{1}{\sqrt{1+e^{2x}}}$.
Result
The solution to the integral is: