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

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

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integral from 0 to 1 of 2xarctan(x)

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

\frac{π - 2}{2}

Decimal Notation

0.57079 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{π - 2}{4}

= 2 \cdot \frac{π - 2}{4}

Simplify

= \frac{π - 2}{2}

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\int_{0}^{4} (\frac{1}{9}) d x

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