What is the integral from 0 to 1/9 of yarctan(9y) ?

The integral from 0 to 1/9 of yarctan(9y) is (2pi-4)/(648)

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integral from 0 to 1/9 of yarctan(9y)

\int_{0}^{\frac{1}{9}} y arctan (9 y) d y

\frac{2 π - 4}{648}

Decimal Notation

0.00352 \dots

Solution steps

\int_{0}^{\frac{1}{9}} y arctan (9 y) d y

Apply Integration By Parts

: [\frac{1}{2} (y^{2} arctan (9 y) - 2 \cdot \int \frac{9 y ^{2}}{2 ( 81 y ^{2} + 1 )} d y)]_{0}^{\frac{1}{9}}

= [\frac{1}{2} (y^{2} arctan (9 y) - 2 \cdot \int \frac{9 y ^{2}}{2 ( 81 y ^{2} + 1 )} d y)]_{0}^{\frac{1}{9}}

\int \frac{9 y ^{2}}{2 ( 81 y ^{2} + 1 )} d y = \frac{1}{162} (- arctan (9 y) + 9 y)

= [\frac{1}{2} (y^{2} arctan (9 y) - 2 \cdot \frac{1}{162} (- arctan (9 y) + 9 y))]_{0}^{\frac{1}{9}}

2 \cdot \frac{1}{162} (- arctan (9 y) + 9 y) = \frac{1}{81} (- arctan (9 y) + 9 y)

= [\frac{1}{2} (y^{2} arctan (9 y) - \frac{1}{81} (- arctan (9 y) + 9 y))]_{0}^{\frac{1}{9}}

Compute the boundaries:

\frac{2 π - 4}{648}

= \frac{2 π - 4}{648}

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What is the integral from 0 to 1/9 of yarctan(9y) ?
The integral from 0 to 1/9 of yarctan(9y) is (2pi-4)/(648)