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

解答

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

- x + \frac{1}{2} (x + 2)^{2} - 3 ln ∣ x + 2 ∣ - 2 + C

求解步骤

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

乘开

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

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

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

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

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

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

化简

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

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

解答补常数

= - x + \frac{1}{2} (x + 2)^{2} - 3 ln ∣ x + 2 ∣ - 2 + C

\int_{0}^{3} (8 - 2 x) d x

f (x) = \frac{6}{x ^{4}}

\int e^{- \frac{3}{2} x} d x

a re a 7 cos (π x), 8 x^{2} - 2, [- 0.5, 0.5]

x \to - 1 lim (\frac{3 x ^{2} - 9 x - 12}{3 x + 3})