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受欢迎的微积分 >

integral of ((8+3x-x^2))/((2x+3)(x+2)^2)

解答

\int \frac{( 8 + 3 x - x ^{2} )}{( 2 x + 3 ) ( x + 2 ) ^{2}} d x

解答

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

求解步骤

\int \frac{8 + 3 x - x ^{2}}{( 2 x + 3 ) ( x + 2 ) ^{2}} d x

使用换元积分法

= \int \frac{- u ^{2} + 12 u + 5}{2 u ( u + 1 ) ^{2}} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{- u ^{2} + 12 u + 5}{u ( u + 1 ) ^{2}} d u

乘开

\frac{- u ^{2} + 12 u + 5}{u ( u + 1 ) ^{2}} : - \frac{u ^{2}}{u ( u + 1 ) ^{2}} + \frac{12 u}{u ( u + 1 ) ^{2}} + \frac{5}{u ( u + 1 ) ^{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

= \frac{1}{2} (- \int \frac{u ^{2}}{u ( u + 1 ) ^{2}} d u + \int \frac{12 u}{u ( u + 1 ) ^{2}} d u + \int \frac{5}{u ( u + 1 ) ^{2}} d u)

\int \frac{u ^{2}}{u ( u + 1 ) ^{2}} d u = ln ∣ u + 1 ∣ + \frac{1}{u + 1}

\int \frac{12 u}{u ( u + 1 ) ^{2}} d u = - \frac{12}{u + 1}

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

= \frac{1}{2} (- (ln ∣ u + 1 ∣ + \frac{1}{u + 1}) - \frac{12}{u + 1} + 5 (ln ∣ u ∣ - ln ∣ u + 1 ∣ + \frac{1}{u + 1}))

u = 2 x + 3

代回

= \frac{1}{2} (- (ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}) - \frac{12}{2 x + 3 + 1} + 5 (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}))

化简

\frac{1}{2} (- (ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1}) - \frac{12}{2 x + 3 + 1} + 5 (ln ∣ 2 x + 3 ∣ - ln ∣ 2 x + 3 + 1 ∣ + \frac{1}{2 x + 3 + 1})) : \frac{( x + 2 ) ( 5 ln ∣ 2 x + 3 ∣ - 6 ln ( 2 ∣ x + 2 ∣ ) ) - 4}{2 ( x + 2 )}

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

解答补常数

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

作图

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