integral of (x^2+2x-1)/((x+1)(x+2)(x+3))

解答

\int \frac{x ^{2} + 2 x - 1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x

- ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣ + C

求解步骤

\int \frac{x ^{2} + 2 x - 1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x

乘开

\frac{x ^{2} + 2 x - 1}{( x + 1 ) ( x + 2 ) ( x + 3 )} : \frac{x ^{2}}{( x + 1 ) ( x + 2 ) ( x + 3 )} + \frac{2 x}{( x + 1 ) ( x + 2 ) ( x + 3 )} - \frac{1}{( x + 1 ) ( x + 2 ) ( x + 3 )}

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 \frac{x ^{2}}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x + \int \frac{2 x}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x - \int \frac{1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x

\int \frac{x ^{2}}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x = \frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣

\int \frac{2 x}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x = - ln ∣ x + 1 ∣ + 4 ln ∣ x + 2 ∣ - 3 ln ∣ x + 3 ∣

\int \frac{1}{( x + 1 ) ( x + 2 ) ( x + 3 )} d x = \frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣

= \frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + 4 ln ∣ x + 2 ∣ - 3 ln ∣ x + 3 ∣ - (\frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣)

化简

\frac{1}{2} ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + \frac{9}{2} ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + 4 ln ∣ x + 2 ∣ - 3 ln ∣ x + 3 ∣ - (\frac{1}{2} ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ + \frac{1}{2} ln ∣ x + 3 ∣) : - ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣

= - ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣

解答补常数

= - ln ∣ x + 1 ∣ + ln ∣ x + 3 ∣ + ln ∣ x + 2 ∣ + C

y^{^{'}} - y = x, y (0) = 1

x \to - 2 lim (x^{3} + 8)

\int x (x - 1)^{4} d x

\int \frac{9}{x ^{2} x ^{2} + 16} d x

d er i v a t i v e tan^{2} (5 θ)