What is the integral from 1 to sqrt(3 of)9arctan(1x) ?

The integral from 1 to sqrt(3 of)9arctan(1x) is (12sqrt(3)pi-9pi-18ln(2))/4

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integral from 1 to sqrt(3 of)9arctan(1x)

\int_{1}^{3} 9 arctan (1 x) d x

\frac{12 3 π - 9 π - 18 ln ( 2 )}{4}

Decimal

6.13644 \dots

Solution steps

\int_{1}^{3} 9 arctan (1 \cdot x) d x

Take the constant out:

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

= 9 \cdot \int_{1}^{3} arctan (1 \cdot x) d x

Simplify

= 9 \cdot \int_{1}^{3} arctan (x) d x

Apply Integration By Parts

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

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

\int \frac{x}{x ^{2} + 1} d x = \frac{1}{2} ln x^{2} + 1

= 9 [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{1}^{3}

Simplify

9 [x arctan (x) - \frac{1}{2} ln x^{2} + 1]_{1}^{3} : 9 [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{1}^{3}

= 9 [\frac{1}{2} (2 x arctan (x) - ln x^{2} + 1)]_{1}^{3}

Compute the boundaries:

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

= 9 \cdot \frac{- 2 3 ln ( 2 ) + 4 π - 3 π}{4 3}

Simplify

= \frac{12 3 π - 9 π - 18 ln ( 2 )}{4}

f (x) = x^{3} 1 - x^{2}

y = 1 + x, y = \frac{( x + 3 )}{3}

\int sec^{3} (9 x) d x

f (x) = (ln (x))^{3}, at x = 3

\int \frac{( 5 x + 1 )}{( 2 x + 1 ) ( x - 1 )} d x

What is the integral from 1 to sqrt(3 of)9arctan(1x) ?
The integral from 1 to sqrt(3 of)9arctan(1x) is (12sqrt(3)pi-9pi-18ln(2))/4