integral of (2x^2-2x-1)/(x^3+x^2)

解

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

\frac{2 x}{x + 1} + 3 ln ∣ x + 1 ∣ + \frac{2}{x + 1} - ln ∣ x ∣ + \frac{1}{x} + C

解答ステップ

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

拡張

\frac{2 x ^{2} - 2 x - 1}{x ^{3} + x ^{2}} : \frac{2 x ^{2}}{x ^{3} + x ^{2}} - \frac{2 x}{x ^{3} + x ^{2}} - \frac{1}{x ^{3} + x ^{2}}

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

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

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

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

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

簡素化

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

= \frac{2 x}{x + 1} + 3 ln ∣ x + 1 ∣ + \frac{2}{x + 1} - ln ∣ x ∣ + \frac{1}{x}

定数を解答に追加する

= \frac{2 x}{x + 1} + 3 ln ∣ x + 1 ∣ + \frac{2}{x + 1} - ln ∣ x ∣ + \frac{1}{x} + C