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

The integral from 0 to 1 of x/(1+x^4) is pi/8

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

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

\frac{π}{8}

Decimal Notation

0.39269 \dots

Solution steps

\int_{0}^{1} \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}^{1} \frac{x}{1 + x ^{4}} d x = \frac{π}{8} - 0

= \frac{π}{8} - 0

Simplify

= \frac{π}{8}

\frac{d y}{d x} = x y + 2 x + y + 2

y = 3 x, [3, 7]

\int 34 arctan (x) d x

x \to 2 lim ((x + 2) \cdot (2 x - 1))

3 + \frac{1}{5} sin (3 π t)

What is the integral from 0 to 1 of x/(1+x^4) ?
The integral from 0 to 1 of x/(1+x^4) is pi/8