What is the integral from 0 to 1/2 of xarctan(2x) ?

The integral from 0 to 1/2 of xarctan(2x) is (2pi-4)/(32)

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integral from 0 to 1/2 of xarctan(2x)

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

\frac{2 π - 4}{32}

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{x ^{2}}{4 x ^{2} + 1} d x = \frac{1}{8} (- arctan (2 x) + 2 x)

= [\frac{1}{2} (x^{2} arctan (2 x) - 2 \cdot \frac{1}{8} (- arctan (2 x) + 2 x))]_{0}^{\frac{1}{2}}

2 \cdot \frac{1}{8} (- arctan (2 x) + 2 x) = \frac{1}{4} (- arctan (2 x) + 2 x)

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

Compute the boundaries:

\frac{2 π - 4}{32}

= \frac{2 π - 4}{32}

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What is the integral from 0 to 1/2 of xarctan(2x) ?
The integral from 0 to 1/2 of xarctan(2x) is (2pi-4)/(32)