What is the integral from 8 to 11 of 2arccot(x) ?

The integral from 8 to 11 of 2arccot(x) is 22arccot(11)-16arccot(8)+ln(122)-ln(65)

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integral from 8 to 11 of 2arccot(x)

\int_{8}^{11} 2 arccot (x) d x

22 arccot (11) - 16 arccot (8) + ln (122) - ln (65)

Decimal Notation

0.63447 \dots

Solution steps

\int_{8}^{11} 2 arccot (x) d x

Take the constant out:

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

= 2 \cdot \int_{8}^{11} arccot (x) d x

Apply Integration By Parts

: [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{8}^{11}

= 2 [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{8}^{11}

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

= 2 [x arccot (x) - (- \frac{1}{2} ln x^{2} + 1)]_{8}^{11}

Simplify

= 2 [x arccot (x) + \frac{1}{2} ln x^{2} + 1]_{8}^{11}

Compute the boundaries:

11 arccot (11) - 8 arccot (8) + \frac{ln ( 122 ) - ln ( 65 )}{2}

= 2 (11 arccot (11) - 8 arccot (8) + \frac{ln ( 122 ) - ln ( 65 )}{2})

Simplify

= 22 arccot (11) - 16 arccot (8) + ln (122) - ln (65)

t^{2} e^{- 2 t} sin (2 t)

\frac{\partial}{\partial x} (2 x + y)

7 y^{^{'}} - 14 y = 0

\int tan (x) d x

\frac{- 3}{2 x}

What is the integral from 8 to 11 of 2arccot(x) ?
The integral from 8 to 11 of 2arccot(x) is 22arccot(11)-16arccot(8)+ln(122)-ln(65)