integral from 1 to 5 of (x^2+2)/(6x-x^2)

解答

\int_{1}^{5} \frac{x ^{2} + 2}{6 x - x ^{2}} d x

6 ln (5) + \frac{2 ln ( 5 )}{3} - 4

十进制

6.72958 \dots

求解步骤

\int_{1}^{5} \frac{x ^{2} + 2}{6 x - x ^{2}} d x

\frac{x ^{2} + 2}{6 x - x ^{2}} = \frac{6 x + 2}{6 x - x ^{2}} - 1

= \int_{1}^{5} \frac{6 x + 2}{6 x - x ^{2}} - 1 d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{5} \frac{6 x + 2}{6 x - x ^{2}} d x - \int_{1}^{5} 1 d x

\int_{1}^{5} \frac{6 x + 2}{6 x - x ^{2}} d x = 6 ln (5) + \frac{2 ln ( 5 )}{3}

\int_{1}^{5} 1 d x = 4

= 6 ln (5) + \frac{2 ln ( 5 )}{3} - 4

a re a y^{2} - 2 x = 2, x - y = 3

\int \frac{2}{x - 1} d x

x \to 3 π + lim (csc (x))

f (x) = - \frac{5}{x ^{2}}

y^{^{''}} + 2 y^{^{'}} + 37 y = 5 e^{- t} cos (6 t)