Calculate the integral: integral from 0 to 1 dx/(e^x+e^-x)

Calculate the integral: $\int_0^1 \frac{dx}{e^x+e^{-x}}$.

We will simplify the integrand using hyperbolic functions, then find the antiderivative.

Step 1: Recognize the hyperbolic cosine

Recall that $\cosh(x) = \frac{e^x + e^{-x}}{2}$, so $e^x+e^{-x}=2\cosh(x)$. The integrand becomes:

\[\frac{1}{e^x+e^{-x}} = \frac{1}{2\cosh(x)}\]

Step 2: Rewrite the integral

\[\int_0^1 \frac{dx}{e^x+e^{-x}} = \frac{1}{2}\int_0^1 \frac{1}{\cosh(x)} \, dx = \frac{1}{2}\int_0^1 \text{sech}(x) \, dx\]

Step 3: Find the antiderivative of $\text{sech}(x)$

We know that $\int \text{sech}(x) \, dx = \arctan(\sinh(x)) + C$ (equivalently $2\arctan(e^x)+C'$, or $\arctan(\tanh(x/2))+C''$). We'll use the first form.

Step 4: Apply the Newton-Leibniz formula

\[\int_0^1 \frac{dx}{e^x+e^{-x}} = \frac{1}{2}\left[\arctan(\sinh(x))\right]_0^1\]

Step 5: Evaluate at the upper limit $x = 1$

\[\sinh(1) = \frac{e - e^{-1}}{2} = \frac{e^2-1}{2e}, \qquad F(1) = \arctan(\sinh(1)) = \arctan\left(\frac{e^2-1}{2e}\right)\]

Step 6: Evaluate at the lower limit $x = 0$

\[F(0) = \arctan(\sinh(0)) = \arctan(0) = 0\]

Step 7: Calculate the definite integral

\[\int_0^1 \frac{dx}{e^x+e^{-x}} = \frac{1}{2}\arctan(\sinh(1)) = \frac{1}{2}\arctan\left(\frac{e^2-1}{2e}\right)\]

Numerical calculation:

Using $e \approx 2.71828$: $\sinh(1)\approx 1.17520$, so $\arctan(1.17520)\approx 0.86750$ radians, and

\[\int_0^1 \frac{dx}{e^x+e^{-x}} \approx \frac{1}{2}(0.86750) \approx 0.43375\]

Alternative method using substitution:

Multiplying numerator and denominator by $e^x$: $\frac{1}{e^x+e^{-x}} = \frac{e^x}{e^{2x}+1}$. Let $u=e^x$, so $du=e^x\,dx$, i.e. $dx=\frac{du}{u}$:

\[\frac{e^x}{e^{2x}+1} \, dx = \frac{u}{u^2+1} \cdot \frac{du}{u} = \frac{du}{u^2+1}\]

When $x=0$: $u=1$. When $x=1$: $u=e$.

\[\int_0^1 \frac{dx}{e^x+e^{-x}} = \int_1^e \frac{du}{u^2+1} = \left[\arctan(u)\right]_1^e = \arctan(e) - \arctan(1) = \arctan(e) - \frac{\pi}{4}\]

Verification:

Using $\arctan(e)\approx 1.21828$ and $\frac\pi4\approx 0.78540$: $\arctan(e)-\frac\pi4 \approx 0.43288$, matching our previous calculation.

Final answer:

\[\boxed{\int_0^1 \frac{dx}{e^x+e^{-x}} = \arctan(e) - \frac{\pi}{4} = \frac{1}{2}\arctan(\sinh(1)) \approx 0.4329}\]