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受欢迎的微积分 >

integral of (7(4-x^2))/((x-4)(x-2)(x+3))

解答

\int \frac{7 ( 4 - x ^{2} )}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x

解答

- 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣ + C

求解步骤

\int \frac{7 ( 4 - x ^{2} )}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x

提出常数:

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

= 7 \cdot \int \frac{4 - x ^{2}}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x

使用换元积分法

= 7 \cdot \int - \frac{u - 1}{u ( u - 7 )} d u

提出常数:

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

= 7 (- \int \frac{u - 1}{u ( u - 7 )} d u)

乘开

\frac{u - 1}{u ( u - 7 )} : \frac{u}{u ( u - 7 )} - \frac{1}{u ( u - 7 )}

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

= 7 (- (\int \frac{u}{u ( u - 7 )} d u - \int \frac{1}{u ( u - 7 )} d u))

\int \frac{u}{u ( u - 7 )} d u = ln ∣ u - 7 ∣

\int \frac{1}{u ( u - 7 )} d u = - \frac{1}{7} ln ∣ u ∣ + \frac{1}{7} ln ∣ u - 7 ∣

= 7 (- (ln ∣ u - 7 ∣ - (- \frac{1}{7} ln ∣ u ∣ + \frac{1}{7} ln ∣ u - 7 ∣)))

u = x + 3

代回

= 7 (- (ln ∣ x + 3 - 7 ∣ - (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x + 3 - 7 ∣)))

化简

7 (- (ln ∣ x + 3 - 7 ∣ - (- \frac{1}{7} ln ∣ x + 3 ∣ + \frac{1}{7} ln ∣ x + 3 - 7 ∣))) : - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣

= - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣

解答补常数

= - 6 ln ∣ x - 4 ∣ - ln ∣ x + 3 ∣ + C

作图

流行的例子

integral of 1/5 sin(5x)

\int \frac{1}{5} sin (5 x) d x

\frac{d}{d t} (- 4 t^{3})

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\int - \frac{1}{8} sin (2 x) d x

limit as x approaches 1 of log_{4}(x)

x \to 1 lim (lo g_{4} (x))

limit as x approaches 8+of 1/(x-8)-1/(sqrt(x-8))

x \to 8 + lim (\frac{1}{x - 8} - \frac{1}{x - 8})