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

解答

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

- \frac{3}{2} arctan (\frac{x}{2}) + \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣ + C

求解步骤

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

乘开

\frac{4 x ^{2} - 3 x + 4}{x ( x ^{2} + 4 )} : \frac{4 x ^{2}}{x ( x ^{2} + 4 )} - \frac{3 x}{x ( x ^{2} + 4 )} + \frac{4}{x ( x ^{2} + 4 )}

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

\int \frac{4 x ^{2}}{x ( x ^{2} + 4 )} d x = 2 ln x^{2} + 4

\int \frac{3 x}{x ( x ^{2} + 4 )} d x = \frac{3}{2} arctan (\frac{x}{2})

\int \frac{4}{x ( x ^{2} + 4 )} d x = ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

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

化简

2 ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) + ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4 : - \frac{3}{2} arctan (\frac{x}{2}) + \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣

= - \frac{3}{2} arctan (\frac{x}{2}) + \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣

解答补常数

= - \frac{3}{2} arctan (\frac{x}{2}) + \frac{3}{2} ln x^{2} + 4 + ln ∣ x ∣ + C

a re a y = \frac{1}{2} x^{2}, \frac{1}{x ^{2} + 1}

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

\frac{\partial}{\partial y} (3 x^{2} + 2 x y)

\frac{d}{d a} (sin^{3} (a) + cos^{3} (a))

t an g e n t f (x) = \frac{1}{5 + 2 x}, (2, \frac{1}{9})