integral of (2x^2-2x-8)/(x^2(x-4))

解答

\int \frac{2 x ^{2} - 2 x - 8}{x ^{2} ( x - 4 )} d x

ln ∣ x ∣ + ln ∣ x - 4 ∣ - \frac{2}{x} + C

求解步骤

\int \frac{2 x ^{2} - 2 x - 8}{x ^{2} ( x - 4 )} d x

乘开

\frac{2 x ^{2} - 2 x - 8}{x ^{2} ( x - 4 )} : \frac{2 x ^{2}}{x ^{2} ( x - 4 )} - \frac{2 x}{x ^{2} ( x - 4 )} - \frac{8}{x ^{2} ( x - 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{2 x ^{2}}{x ^{2} ( x - 4 )} d x - \int \frac{2 x}{x ^{2} ( x - 4 )} d x - \int \frac{8}{x ^{2} ( x - 4 )} d x

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

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

\int \frac{8}{x ^{2} ( x - 4 )} d x = 8 (- \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣)

= 2 ln ∣ x - 4 ∣ - 2 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣) - 8 (- \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣)

化简

2 ln ∣ x - 4 ∣ - 2 (- \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣) - 8 (- \frac{1}{16} ln ∣ x ∣ + \frac{1}{4 x} + \frac{1}{16} ln ∣ x - 4 ∣) : ln ∣ x ∣ + ln ∣ x - 4 ∣ - \frac{2}{x}

= ln ∣ x ∣ + ln ∣ x - 4 ∣ - \frac{2}{x}

解答补常数

= ln ∣ x ∣ + ln ∣ x - 4 ∣ - \frac{2}{x} + C

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

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

\frac{d}{d x} (\frac{x ^{2} + 100}{x})

x \to 2 lim (ln (2))

\frac{d}{d x} (- \frac{1}{x ^{4}} - \frac{5}{x ^{2}})