What is the integral from 5 to 6 of 12arccot(x) ?

The integral from 5 to 6 of 12arccot(x) is 12(6arccot(6)-5arccot(5)+(ln(37)-ln(26))/2)

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integral from 5 to 6 of 12arccot(x)

\int_{5}^{6} 12 arccot (x) d x

12 (6 arccot (6) - 5 arccot (5) + \frac{ln ( 37 ) - ln ( 26 )}{2})

Decimal Notation

2.16389 \dots

Solution steps

\int_{5}^{6} 12 arccot (x) d x

Take the constant out:

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

= 12 \cdot \int_{5}^{6} arccot (x) d x

Apply Integration By Parts

: [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{5}^{6}

= 12 [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{5}^{6}

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

= 12 [x arccot (x) - (- \frac{1}{2} ln x^{2} + 1)]_{5}^{6}

Simplify

= 12 [x arccot (x) + \frac{1}{2} ln x^{2} + 1]_{5}^{6}

Compute the boundaries:

6 arccot (6) - 5 arccot (5) + \frac{ln ( 37 ) - ln ( 26 )}{2}

= 12 (6 arccot (6) - 5 arccot (5) + \frac{ln ( 37 ) - ln ( 26 )}{2})

(- 5, 3), (5, - 5)

\frac{d}{d x} (x cos (θ))

\int x y e^{- y x} d x

x^{2} + x y - y^{2} = - 11, (1, 4)

What is the integral from 5 to 6 of 12arccot(x) ?
The integral from 5 to 6 of 12arccot(x) is 12(6arccot(6)-5arccot(5)+(ln(37)-ln(26))/2)