What is the integral from 0 to 2 of x*arctan(x/2) ?

The integral from 0 to 2 of x*arctan(x/2) is pi-2

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integral from 0 to 2 of x*arctan(x/2)

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

π - 2

Decimal Notation

1.14159 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} 4 u arctan (u) d u

= \int_{0}^{1} 4 u arctan (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{0}^{1} u arctan (u) d u

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π - 2}{4}

= 4 \cdot \frac{π - 2}{4}

Simplify

= π - 2

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What is the integral from 0 to 2 of x*arctan(x/2) ?
The integral from 0 to 2 of x*arctan(x/2) is pi-2