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

The integral from 1 to 3 of arctan(sqrt(x)) is (5pi-6sqrt(3)+6)/6

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

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

\frac{5 π - 6 3 + 6}{6}

Decimal

1.88594 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{3} arctan (u) \cdot 2 u d u

= \int_{1}^{3} arctan (u) \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int_{1}^{3} arctan (u) u d u

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Compute the boundaries:

\frac{5 π - 6 3 + 6}{12}

= 2 \cdot \frac{5 π - 6 3 + 6}{12}

Simplify

= \frac{5 π - 6 3 + 6}{6}

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What is the integral from 1 to 3 of arctan(sqrt(x)) ?
The integral from 1 to 3 of arctan(sqrt(x)) is (5pi-6sqrt(3)+6)/6