integral of (10x^2-9x+20)/(x^3+4x)

解答

\int \frac{10 x ^{2} - 9 x + 20}{x ^{3} + 4 x} d x

- \frac{9}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 5 ln ∣ x ∣ + C

求解步骤

\int \frac{10 x ^{2} - 9 x + 20}{x ^{3} + 4 x} d x

乘开

\frac{10 x ^{2} - 9 x + 20}{x ^{3} + 4 x} : \frac{10 x ^{2}}{x ^{3} + 4 x} - \frac{9 x}{x ^{3} + 4 x} + \frac{20}{x ^{3} + 4 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 \frac{10 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{9 x}{x ^{3} + 4 x} d x + \int \frac{20}{x ^{3} + 4 x} d x

\int \frac{10 x ^{2}}{x ^{3} + 4 x} d x = 5 ln x^{2} + 4

\int \frac{9 x}{x ^{3} + 4 x} d x = \frac{9}{2} arctan (\frac{x}{2})

\int \frac{20}{x ^{3} + 4 x} d x = 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= 5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

化简

5 ln x^{2} + 4 - \frac{9}{2} arctan (\frac{x}{2}) + 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : - \frac{9}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 5 ln ∣ x ∣

= - \frac{9}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 5 ln ∣ x ∣

解答补常数

= - \frac{9}{2} arctan (\frac{x}{2}) + \frac{5}{2} ln x^{2} + 4 + 5 ln ∣ x ∣ + C

x y^{^{'}} + x y - y^{^{'}} - 2 y = 0

\int - 2 tan (2 x) d x

x^{2} y^{^{'}} + x y = 7

x \to 1 + lim (x^{2} + 2 x)

x \to 0 lim (\frac{( sin ( 6 x ) )}{x})