integral of (2x^2-5x+4)/((x-2)^3)

解

\int \frac{2 x ^{2} - 5 x + 4}{( x - 2 ) ^{3}} d x

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

解答ステップ

\int \frac{2 x ^{2} - 5 x + 4}{( x - 2 ) ^{3}} d x

拡張

\frac{2 x ^{2} - 5 x + 4}{( x - 2 ) ^{3}} : \frac{2 x ^{2}}{( x - 2 ) ^{3}} - \frac{5 x}{( x - 2 ) ^{3}} + \frac{4}{( x - 2 ) ^{3}}

= \int \frac{2 x ^{2}}{( x - 2 ) ^{3}} - \frac{5 x}{( x - 2 ) ^{3}} + \frac{4}{( x - 2 ) ^{3}} d 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 - 2 ) ^{3}} d x - \int \frac{5 x}{( x - 2 ) ^{3}} d x + \int \frac{4}{( x - 2 ) ^{3}} d x

\int \frac{2 x ^{2}}{( x - 2 ) ^{3}} d x = - \frac{x ^{2}}{( x - 2 ) ^{2}} + 2 ln ∣ x - 2 ∣ - \frac{4}{x - 2}

\int \frac{5 x}{( x - 2 ) ^{3}} d x = 5 (- \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}})

\int \frac{4}{( x - 2 ) ^{3}} d x = - \frac{2}{( x - 2 ) ^{2}}

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

簡素化

- \frac{x ^{2}}{( x - 2 ) ^{2}} + 2 ln ∣ x - 2 ∣ - \frac{4}{x - 2} - 5 (- \frac{1}{x - 2} - \frac{1}{( x - 2 ) ^{2}}) - \frac{2}{( x - 2 ) ^{2}} : - \frac{x ^{2}}{( x - 2 ) ^{2}} + 2 ln ∣ x - 2 ∣ + \frac{1}{x - 2} + \frac{3}{( x - 2 ) ^{2}}

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

定数を解答に追加する

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