What is the integral from 0 to 0.1 of xarctan(3x) ?

The integral from 0 to 0.1 of xarctan(3x) is 0.00098…

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integral from 0 to 0.1 of xarctan(3x)

\int_{0}^{0.1} x arctan (3 x) d x

0.00098 \dots

Solution steps

\int_{0}^{0.1} x arctan (3 x) d x

Apply Integration By Parts

: [\frac{1}{2} (x^{2} arctan (3 x) - 2 \cdot \int \frac{3 x ^{2}}{2 ( 9 x ^{2} + 1 )} d x)]_{0}^{0.1}

= [\frac{1}{2} (x^{2} arctan (3 x) - 2 \cdot \int \frac{3 x ^{2}}{2 ( 9 x ^{2} + 1 )} d x)]_{0}^{0.1}

\int \frac{3 x ^{2}}{2 ( 9 x ^{2} + 1 )} d x = \frac{1}{18} (- arctan (3 x) + 3 x)

= [\frac{1}{2} (x^{2} arctan (3 x) - 2 \cdot \frac{1}{18} (- arctan (3 x) + 3 x))]_{0}^{0.1}

Simplify

[\frac{1}{2} (x^{2} arctan (3 x) - 2 \cdot \frac{1}{18} (- arctan (3 x) + 3 x))]_{0}^{0.1} : [0.5 (x^{2} arctan (3 x) - 0.11111 \dots (- arctan (3 x) + 3 x))]_{0}^{0.1}

= [0.5 (x^{2} arctan (3 x) - 0.11111 \dots (- arctan (3 x) + 3 x))]_{0}^{0.1}

Compute the boundaries:

0.00098 \dots

= 0.00098 \dots

\frac{d}{d x} (1. 5^{1.7 x})

(- 5, 1), (- 4, - 9)

\frac{d}{d x} (200 ln (\frac{x}{40}))

\int (3 t + 1) (t - 2) d t

n \to \infty lim (n!)

What is the integral from 0 to 0.1 of xarctan(3x) ?
The integral from 0 to 0.1 of xarctan(3x) is 0.00098…