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

解

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

- 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{2} + 2 x + 1}{( x - 2 ) ^{2} ( x - 1 )} : - \frac{3}{x - 2} + \frac{9}{( x - 2 ) ^{2}} + \frac{4}{x - 1}

= \int - \frac{3}{x - 2} + \frac{9}{( x - 2 ) ^{2}} + \frac{4}{x - 1} 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{3}{x - 2} d x + \int \frac{9}{( x - 2 ) ^{2}} d x + \int \frac{4}{x - 1} d x

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

\int \frac{9}{( x - 2 ) ^{2}} d x = - \frac{9}{x - 2}

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= - 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣

定数を解答に追加する

= - 3 ln ∣ x - 2 ∣ - \frac{9}{x - 2} + 4 ln ∣ x - 1 ∣ + C