What is the integral from 0 to 1 of (x^2)/(1+x^6) ?

The integral from 0 to 1 of (x^2)/(1+x^6) is pi/(12)

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integral from 0 to 1 of (x^2)/(1+x^6)

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{6}} d x

\frac{π}{12}

Decimal

0.26179 \dots

Solution steps

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{6}} d x

Compute the indefinite integral:

\int \frac{x ^{2}}{1 + x ^{6}} d x = \frac{1}{3} arctan (x^{3}) + C

Compute the boundaries:

\int_{0}^{1} \frac{x ^{2}}{1 + x ^{6}} d x = \frac{π}{12} - 0

= \frac{π}{12} - 0

Simplify

= \frac{π}{12}

\int e^{- 2 l n (x)} d x

\frac{\partial}{\partial x} (e^{x z} + x y)

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

\frac{d y}{d x} = x y + 1

\frac{d}{d x} (\frac{e ^{x}}{1 + x})

What is the integral from 0 to 1 of (x^2)/(1+x^6) ?
The integral from 0 to 1 of (x^2)/(1+x^6) is pi/(12)