integral of (x+9)/(2x^3-18x)

解答

\int \frac{x + 9}{2 x ^{3} - 18 x} d x

\frac{1}{6} (ln ∣ x + 3 ∣ + 2 ln ∣ x - 3 ∣ - 3 ln ∣ x ∣) + C

求解步骤

\int \frac{x + 9}{2 x ^{3} - 18 x} d x

分解

\frac{x + 9}{2 x ^{3} - 18 x}

= \int \frac{x + 9}{2 x ( x + 3 ) ( x - 3 )} d x

提出常数:

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

= \frac{1}{2} \cdot \int \frac{x + 9}{x ( x + 3 ) ( x - 3 )} d x

乘开

\frac{x + 9}{x ( x + 3 ) ( x - 3 )} : \frac{x}{x ( x + 3 ) ( x - 3 )} + \frac{9}{x ( x + 3 ) ( 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

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

\int \frac{x}{x ( x + 3 ) ( x - 3 )} d x = - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣

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

= \frac{1}{2} (- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + \frac{1}{2} ln ∣ x + 3 ∣ - ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 3 ∣)

化简

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

= \frac{1}{6} (ln ∣ x + 3 ∣ + 2 ln ∣ x - 3 ∣ - 3 ln ∣ x ∣)

解答补常数

= \frac{1}{6} (ln ∣ x + 3 ∣ + 2 ln ∣ x - 3 ∣ - 3 ln ∣ x ∣) + C

2 \frac{d y}{d x} + 18 y = 9

\int 5 sin^{2} (3 x) d x

\int_{0}^{1} \frac{1}{x ^{0.2}} d x

x \to 0 - lim (x (x - 1))

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