What is the integral from 1 to 2 of x/(1+x^4) ?

The integral from 1 to 2 of x/(1+x^4) is 1/2 arctan(4)-pi/8

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integral from 1 to 2 of x/(1+x^4)

\int_{1}^{2} \frac{x}{1 + x ^{4}} d x

\frac{1}{2} arctan (4) - \frac{π}{8}

Decimal Notation

0.27020 \dots

Solution steps

\int_{1}^{2} \frac{x}{1 + x ^{4}} d x

Compute the indefinite integral:

\int \frac{x}{1 + x ^{4}} d x = \frac{1}{2} arctan (x^{2}) + C

Compute the boundaries:

\int_{1}^{2} \frac{x}{1 + x ^{4}} d x = \frac{1}{2} arctan (4) - \frac{π}{8}

= \frac{1}{2} arctan (4) - \frac{π}{8}

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What is the integral from 1 to 2 of x/(1+x^4) ?
The integral from 1 to 2 of x/(1+x^4) is 1/2 arctan(4)-pi/8