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

解

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

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

解答ステップ

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

拡張

\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 )}

総和規則を適用する:

\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 ∣)

簡素化

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}

定数を解答に追加する

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