integral from 6 to 11 of 10arccot(x)

解答

\int_{6}^{11} 10 arccot (x) d x

10 (11 arccot (11) - 6 arccot (6) + \frac{ln ( 122 ) - ln ( 37 )}{2})

十进制

6.02918 \dots

求解步骤

\int_{6}^{11} 10 arccot (x) d x

提出常数:

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

= 10 \cdot \int_{6}^{11} arccot (x) d x

使用分布积分法

= 10 [x arccot (x) - \int - \frac{x}{x ^{2} + 1} d x]_{6}^{11}

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

= 10 [x arccot (x) - (- \frac{1}{2} ln x^{2} + 1)]_{6}^{11}

化简

= 10 [x arccot (x) + \frac{1}{2} ln x^{2} + 1]_{6}^{11}

带入上下限计算后得:

11 arccot (11) - 6 arccot (6) + \frac{ln ( 122 ) - ln ( 37 )}{2}

= 10 (11 arccot (11) - 6 arccot (6) + \frac{ln ( 122 ) - ln ( 37 )}{2})

d er i v a t i v e f (x) = sin (sin (8 e^{x}))

\int \frac{1}{u ^{2} - a ^{2}} d u

d er i v a t i v e x^{8} (x - 4)^{7}

\frac{d}{d x} (cos (2^{3 x^{2} + 5 x - 6}))

y^{^{'}} = y (x y^{6} + 2)