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

The integral from 0 to 1 of xarctan(1-x) is 1/2-1/2 ln(2)

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integral from 0 to 1 of xarctan(1-x)

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

\frac{1}{2} - \frac{1}{2} ln (2)

Decimal

0.15342 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{1}{2} - \frac{1}{2} ln (2)

= \frac{1}{2} - \frac{1}{2} ln (2)

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