What is the integral from 1 to 3 of sqrt(x)arctan(sqrt(x)) ?

The integral from 1 to 3 of sqrt(x)arctan(sqrt(x)) is (4sqrt(3)pi-pi+2ln(2)-4)/6

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integral from 1 to 3 of sqrt(x)arctan(sqrt(x))

\int_{1}^{3} x arctan (x) d x

\frac{4 3 π - π + 2 ln ( 2 ) - 4}{6}

Decimal Notation

2.66838 \dots

Solution steps

\int_{1}^{3} x arctan (x) d x

Apply Integration By Parts

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

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

\int \frac{x}{3 ( x + 1 )} d x = \frac{1}{3} (x + 1 - ln ∣ x + 1 ∣)

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

Simplify

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

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

Compute the boundaries:

\frac{4 3 π - π + 2 ln ( 2 ) - 4}{6}

= \frac{4 3 π - π + 2 ln ( 2 ) - 4}{6}

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What is the integral from 1 to 3 of sqrt(x)arctan(sqrt(x)) ?
The integral from 1 to 3 of sqrt(x)arctan(sqrt(x)) is (4sqrt(3)pi-pi+2ln(2)-4)/6