What is the integral from 0 to sqrt(3 of)xarctan(x) ?

The integral from 0 to sqrt(3 of)xarctan(x) is (4pi-3sqrt(3))/6

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integral from 0 to sqrt(3 of)xarctan(x)

\int_{0}^{3} x arctan (x) d x

\frac{4 π - 3 3}{6}

Decimal

1.22836 \dots

Solution steps

\int_{0}^{3} 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}^{3}

= [\frac{1}{2} (x^{2} arctan (x) - 2 \cdot \int \frac{x ^{2}}{2 ( x ^{2} + 1 )} d x)]_{0}^{3}

\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}^{3}

Simplify

[\frac{1}{2} (x^{2} arctan (x) - 2 \cdot \frac{1}{2} (- arctan (x) + x))]_{0}^{3} : [\frac{1}{2} (x^{2} arctan (x) + arctan (x) - x)]_{0}^{3}

= [\frac{1}{2} (x^{2} arctan (x) + arctan (x) - x)]_{0}^{3}

Compute the boundaries:

\frac{4 π - 3 3}{6}

= \frac{4 π - 3 3}{6}

\int_{0}^{2 π} x cos^{2} (x) d x

\int_{2}^{5} \frac{3 x ^{2} + 1}{x} d x

\int_{2}^{8} \frac{1}{x} d x

\int_{0}^{1} x^{6} d x

\int_{0}^{2 π} - sin^{2} (t) d t

What is the integral from 0 to sqrt(3 of)xarctan(x) ?
The integral from 0 to sqrt(3 of)xarctan(x) is (4pi-3sqrt(3))/6