integral of (2x^2-4x+3)/((x-1)^2)

解

\int \frac{2 x ^{2} - 4 x + 3}{( x - 1 ) ^{2}} d x

2 x - \frac{1}{x - 1} - 2 + C

解答ステップ

\int \frac{2 x ^{2} - 4 x + 3}{( x - 1 ) ^{2}} d x

拡張

\frac{2 x ^{2} - 4 x + 3}{( x - 1 ) ^{2}} : \frac{2 x ^{2}}{( x - 1 ) ^{2}} - \frac{4 x}{( x - 1 ) ^{2}} + \frac{3}{( x - 1 ) ^{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 - 1 ) ^{2}} d x - \int \frac{4 x}{( x - 1 ) ^{2}} d x + \int \frac{3}{( x - 1 ) ^{2}} d x

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

\int \frac{4 x}{( x - 1 ) ^{2}} d x = 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

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

= 2 (x - 1 + 2 ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - 4 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - \frac{3}{x - 1}

簡素化

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

= 2 x - \frac{1}{x - 1} - 2

定数を解答に追加する

= 2 x - \frac{1}{x - 1} - 2 + C