What is the integral from 0 to pi/4 of xarctan(x) ?

The integral from 0 to pi/4 of xarctan(x) is (arctan(pi/4)(pi^2+16)-4pi)/(32)

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integral from 0 to pi/4 of xarctan(x)

\int_{0}^{\frac{π}{4}} x arctan (x) d x

\frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

Decimal

0.14552 \dots

Solution steps

\int_{0}^{\frac{π}{4}} x arctan (x) d x

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

= \frac{arctan ( \frac{π}{4} ) ( π ^{2} + 16 ) - 4 π}{32}

What is the integral from 0 to pi/4 of xarctan(x) ?
The integral from 0 to pi/4 of xarctan(x) is (arctan(pi/4)(pi^2+16)-4pi)/(32)