integral of 1/(2x^3+2x^2)

解答

\int \frac{1}{2 x ^{3} + 2 x ^{2}} d x

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

求解步骤

\int \frac{1}{2 x ^{3} + 2 x ^{2}} d x

\frac{1}{2 x ^{3} + 2 x ^{2}} = \frac{1}{2} \frac{1}{x ^{3} + x ^{2}}

= \int \frac{1}{2} \cdot \frac{1}{x ^{3} + 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{1}{x ^{3} + x ^{2}} d x

将

\frac{1}{x ^{3} + x ^{2}}

用部份分式展开:

- \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x + 1}

= \frac{1}{2} \cdot \int - \frac{1}{x} + \frac{1}{x ^{2}} + \frac{1}{x + 1} 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{1}{x} d x + \int \frac{1}{x ^{2}} d x + \int \frac{1}{x + 1} d x)

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

解答补常数

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

\frac{d}{d x} (arcsin (3 x^{2}))

d er i v a t i v e - 2 y

t an g e n t f (x) = 5 e^{x} + x, (0, 5)

x \to b lim (\frac{x - b}{x - b})

\frac{\partial}{\partial y} (\frac{1}{y})