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

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

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

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

arctan (e) - \frac{π}{4}

Decimal Notation

0.43288 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

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

Compute the boundaries:

arctan (e) - \frac{π}{4}

= arctan (e) - \frac{π}{4}

\frac{d}{d x} (- 3 ln (3 + 4 x^{2}) + 5 x^{3} - 9)

y = 6

t^{2} \frac{d y}{d t} - t = 1 + y + t y, y (1) = 10

\frac{d}{d x} (6 x^{6})

\int_{- 35}^{0} \frac{1}{1 - x} d x

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