integral of (2x^3-6x^2-10x-6)/(x^2-1)

解

\int \frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} d x

x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1 + C

解答ステップ

\int \frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} d x

拡張

\frac{2 x ^{3} - 6 x ^{2} - 10 x - 6}{x ^{2} - 1} : \frac{2 x ^{3}}{x ^{2} - 1} - \frac{6 x ^{2}}{x ^{2} - 1} - \frac{10 x}{x ^{2} - 1} - \frac{6}{x ^{2} - 1}

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 ^{3}}{x ^{2} - 1} d x - \int \frac{6 x ^{2}}{x ^{2} - 1} d x - \int \frac{10 x}{x ^{2} - 1} d x - \int \frac{6}{x ^{2} - 1} d x

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

\int \frac{6 x ^{2}}{x ^{2} - 1} d x = - 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)

\int \frac{10 x}{x ^{2} - 1} d x = 5 ln x^{2} - 1

\int \frac{6}{x ^{2} - 1} d x = - 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= x^{2} - 1 + ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) - 5 ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣))

簡素化

x^{2} - 1 + ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)) - 5 ln x^{2} - 1 - (- 6 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)) : x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1

= x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1

定数を解答に追加する

= x^{2} - 6 x - 4 ln x^{2} - 1 + 6 ln ∣ x + 1 ∣ - 6 ln ∣ x - 1 ∣ - 1 + C