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

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

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integral from 0 to 2 of x/(x^4+1)

\int_{0}^{2} \frac{x}{x ^{4} + 1} d x

\frac{1}{2} arctan (4)

Decimal Notation

0.66290 \dots

Solution steps

\int_{0}^{2} \frac{x}{x ^{4} + 1} d x

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{1}{2} arctan (4) - 0

Simplify

= \frac{1}{2} arctan (4)

e^{- x^{2}}, 0

\frac{d}{d x} ((x^{4} + 1)^{\frac{1}{2}})

\int 2 z d z

\frac{7 x - 1}{x ^{2} + 9 x}

\int_{- 11}^{0} x 25 - x d x

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