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

0.43288 \dots

Solution steps

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

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\int_{1}^{\infty} 4 x^{- 5} 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