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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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