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

解答

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

\frac{3 x ^{2}}{2} + 3 x + ln ∣ x ∣ + C

求解步骤

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

乘开

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

= \int 3 x + 3 + \frac{1}{x} 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

= \int 3 x d x + \int 3 d x + \int \frac{1}{x} d x

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 3 d x = 3 x

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{3 x ^{2}}{2} + 3 x + ln ∣ x ∣

解答补常数

= \frac{3 x ^{2}}{2} + 3 x + ln ∣ x ∣ + C

s l o p e f (x) = \frac{x ^{2}}{x + 9}

\int \frac{1}{cos ^{2} ( 7 x )} d x

(6 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0

x^{2} \frac{d y}{d x} + 2 x y = 2 x

d er i v a t i v e \frac{x}{6 x + 2}