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

解答

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

\frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣) + C

求解步骤

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

分解

\frac{x + 3}{2 x ^{3} - 8 x}

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

将

\frac{x + 3}{x ( x + 2 ) ( x - 2 )}

用部份分式展开:

- \frac{3}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{5}{8 ( x - 2 )}

= \frac{1}{2} \cdot \int - \frac{3}{4 x} + \frac{1}{8 ( x + 2 )} + \frac{5}{8 ( x - 2 )} d x

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{3}{4 x} d x + \int \frac{1}{8 ( x + 2 )} d x + \int \frac{5}{8 ( x - 2 )} d x)

\int \frac{3}{4 x} d x = \frac{3}{4} ln ∣ x ∣

\int \frac{1}{8 ( x + 2 )} d x = \frac{1}{8} ln ∣ x + 2 ∣

\int \frac{5}{8 ( x - 2 )} d x = \frac{5}{8} ln ∣ x - 2 ∣

= \frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣)

解答补常数

= \frac{1}{2} (- \frac{3}{4} ln ∣ x ∣ + \frac{1}{8} ln ∣ x + 2 ∣ + \frac{5}{8} ln ∣ x - 2 ∣) + C

t an g e n t y = 2 x tan (x), at x = π

x \to 0 lim (\frac{2 x - 1}{x})

\int \frac{sin ( 2 x )}{2 cos ^{2} ( x )} d x

a re a 1 + x - x^{2} - x^{3}, 1 - x, [- 2, 1]

\frac{\partial}{\partial y} (cos (yz))