What is the integral from 0 to 4 of x/(1+x^4) ?

The integral from 0 to 4 of x/(1+x^4) is 1/2 arctan(16)

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integral from 0 to 4 of x/(1+x^4)

\int_{0}^{4} \frac{x}{1 + x ^{4}} d x

\frac{1}{2} arctan (16)

Decimal Notation

0.75418 \dots

Solution steps

\int_{0}^{4} \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_{0}^{4} \frac{x}{1 + x ^{4}} d x = \frac{1}{2} arctan (16) - 0

= \frac{1}{2} arctan (16) - 0

Simplify

= \frac{1}{2} arctan (16)

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What is the integral from 0 to 4 of x/(1+x^4) ?
The integral from 0 to 4 of x/(1+x^4) is 1/2 arctan(16)