integral from 0 to 1 of 1/((x+1)(x+2))

解答

\int_{0}^{1} \frac{1}{( x + 1 ) ( x + 2 )} d x

2 ln (2) - ln (3)

十进制

0.28768 \dots

求解步骤

\int_{0}^{1} \frac{1}{( x + 1 ) ( x + 2 )} d x

将

\frac{1}{( x + 1 ) ( x + 2 )}

用部份分式展开:

\frac{1}{x + 1} - \frac{1}{x + 2}

= \int_{0}^{1} \frac{1}{x + 1} - \frac{1}{x + 2} 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_{0}^{1} \frac{1}{x + 1} d x - \int_{0}^{1} \frac{1}{x + 2} d x

\int_{0}^{1} \frac{1}{x + 1} d x = ln (2)

\int_{0}^{1} \frac{1}{x + 2} d x = ln (3) - ln (2)

= ln (2) - (ln (3) - ln (2))

化简

= 2 ln (2) - ln (3)

t an g e n t f (x) = 17 tan (x), at x = \frac{π}{4}

d er i v a t i v e 5 x^{4} 25 - x^{2}

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

\frac{\partial}{\partial x} (\frac{x}{x ^{6} - y ^{3}})

\frac{d y}{d x} = 4 x + y, y (0) = 9