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

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

Step 1 · Clear the negative exponent

Multiplying numerator and denominator by $e^x$:

$$ I = \int \frac{dx}{e^x+e^{-x}} = \int \frac{e^x\,dx}{e^{2x}+1}. $$

Step 2 · Substitute

Let $v = e^x$. Then $dv = e^x\,dx$, so the integral becomes:

$$ I = \int \frac{dv}{v^2+1}. $$

Step 3 · Apply the standard result

This is a standard integral:

$$ \int \frac{dv}{v^2+1} = \arctan(v) + C. $$

Step 4 · Substitute back

Substituting back $v = e^x$, we obtain:

$$ I = \arctan\left(e^x\right) + C. $$

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $e^x+e^{-x} \geq 2 > 0$ always.
• Using the identity $e^x+e^{-x}=2\cosh x$, the integral can also be written as $\tfrac12\int \operatorname{sech}x\,dx = \tfrac12\arctan(\sinh x)+C$; this differs from $\arctan(e^x)+C$ only by the constant $\arctan(1)=\tfrac\pi4$, so both are valid antiderivatives.
• The result can be checked by differentiating $\arctan(e^x)$ with the chain rule: $\dfrac{e^x}{1+e^{2x}} = \dfrac{1}{e^{-x}+e^x}$, confirming it matches the original integrand.