integral of (x^2-5)/(x+2)

解答

\int \frac{x ^{2} - 5}{x + 2} d x

- 4 x + \frac{1}{2} (x + 2)^{2} - ln ∣ x + 2 ∣ - 12 + C

求解步骤

\int \frac{x ^{2} - 5}{x + 2} d x

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

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

使用换元积分法

= \int \frac{( u + 2 + 5 ) ( u + 2 - 5 )}{u + 4} d u

\frac{( u + 2 + 5 ) ( u + 2 - 5 )}{u + 4} = \frac{u ^{2} + 3 u - 5}{u + 4} + 1

= \int \frac{u ^{2} + 3 u - 5}{u + 4} + 1 d u

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{u ^{2} + 3 u - 5}{u + 4} d u + \int 1 d u

\int \frac{u ^{2} + 3 u - 5}{u + 4} d u = - 5 u + \frac{1}{2} (u + 4)^{2} - ln ∣ u + 4 ∣ - 20

\int 1 d u = u

= - 5 u + \frac{1}{2} (u + 4)^{2} - ln ∣ u + 4 ∣ - 20 + u

u = x - 2

代回

= - 5 (x - 2) + \frac{1}{2} (x - 2 + 4)^{2} - ln ∣ x - 2 + 4 ∣ - 20 + x - 2

化简

- 5 (x - 2) + \frac{1}{2} (x - 2 + 4)^{2} - ln ∣ x - 2 + 4 ∣ - 20 + x - 2 : - 4 x + \frac{1}{2} (x + 2)^{2} - ln ∣ x + 2 ∣ - 12

= - 4 x + \frac{1}{2} (x + 2)^{2} - ln ∣ x + 2 ∣ - 12

解答补常数

= - 4 x + \frac{1}{2} (x + 2)^{2} - ln ∣ x + 2 ∣ - 12 + C