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

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

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

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

arctan (e) - \frac{π}{4}

Decimal

0.43288 \dots

Solution steps

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

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\int_{0}^{1} x^{2} (2 + 2 x^{3})^{3} d x

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