What is the integral from 0 to 1 of (e^x)/(1+e^{2x)} ?

The integral from 0 to 1 of (e^x)/(1+e^{2x)} is arctan(e)-pi/4

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integral from 0 to 1 of (e^x)/(1+e^{2x)}

\int_{0}^{1} \frac{e ^{x}}{1 + e ^{2 x}} d x

arctan (e) - \frac{π}{4}

Decimal

0.43288 \dots

Solution steps

\int_{0}^{1} \frac{e ^{x}}{1 + e ^{2 x}} d x

Apply u-substitution

: \int_{1}^{e} \frac{1}{1 + u ^{2}} d u

= \int_{1}^{e} \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= [arctan (u)]_{1}^{e}

Compute the boundaries:

arctan (e) - \frac{π}{4}

= arctan (e) - \frac{π}{4}

x \to 3 lim (\frac{1}{3} x^{2} - 4 x + 10)

\frac{\partial}{\partial y} (\frac{3 y}{x})

\int \frac{x}{a + b x} d x

\int_{6}^{8} \frac{76}{( x - 6 ) ^{3}} d x

\frac{d y}{d x}, x^{2} y = 1

What is the integral from 0 to 1 of (e^x)/(1+e^{2x)} ?
The integral from 0 to 1 of (e^x)/(1+e^{2x)} is arctan(e)-pi/4