What is the integral from 0 to 1/2 of x^3arctan(x) ?

The integral from 0 to 1/2 of x^3arctan(x) is (-45arctan(1/2)+22)/(192)

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integral from 0 to 1/2 of x^3arctan(x)

\int_{0}^{\frac{1}{2}} x^{3} arctan (x) d x

\frac{- 45 arctan ( \frac{1}{2} ) + 22}{192}

Decimal Notation

0.00591 \dots

Solution steps

\int_{0}^{\frac{1}{2}} x^{3} arctan (x) d x

Apply Integration By Parts

: [\frac{1}{4} (x^{4} arctan (x) - 4 \cdot \int \frac{x ^{4}}{4 ( x ^{2} + 1 )} d x)]_{0}^{\frac{1}{2}}

= [\frac{1}{4} (x^{4} arctan (x) - 4 \cdot \int \frac{x ^{4}}{4 ( x ^{2} + 1 )} d x)]_{0}^{\frac{1}{2}}

\int \frac{x ^{4}}{4 ( x ^{2} + 1 )} d x = \frac{1}{4} (\frac{x ^{3}}{3} - x + arctan (x))

= [\frac{1}{4} (x^{4} arctan (x) - 4 \cdot \frac{1}{4} (\frac{x ^{3}}{3} - x + arctan (x)))]_{0}^{\frac{1}{2}}

Simplify

[\frac{1}{4} (x^{4} arctan (x) - 4 \cdot \frac{1}{4} (\frac{x ^{3}}{3} - x + arctan (x)))]_{0}^{\frac{1}{2}} : [\frac{1}{12} (3 x^{4} arctan (x) - x^{3} + 3 x - 3 arctan (x))]_{0}^{\frac{1}{2}}

= [\frac{1}{12} (3 x^{4} arctan (x) - x^{3} + 3 x - 3 arctan (x))]_{0}^{\frac{1}{2}}

Compute the boundaries:

\frac{- 45 arctan ( \frac{1}{2} ) + 22}{192}

= \frac{- 45 arctan ( \frac{1}{2} ) + 22}{192}

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What is the integral from 0 to 1/2 of x^3arctan(x) ?
The integral from 0 to 1/2 of x^3arctan(x) is (-45arctan(1/2)+22)/(192)