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

解

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

5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3} + C

解答ステップ

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

以下の部分分数を得る:

\frac{2 x + 1}{( x - 2 ) ( x - 3 ) ^{2}} : \frac{5}{x - 2} - \frac{5}{x - 3} + \frac{7}{( x - 3 ) ^{2}}

= \int \frac{5}{x - 2} - \frac{5}{x - 3} + \frac{7}{( x - 3 ) ^{2}} 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{5}{x - 2} d x - \int \frac{5}{x - 3} d x + \int \frac{7}{( x - 3 ) ^{2}} d x

\int \frac{5}{x - 2} d x = 5 ln ∣ x - 2 ∣

\int \frac{5}{x - 3} d x = 5 ln ∣ x - 3 ∣

\int \frac{7}{( x - 3 ) ^{2}} d x = - \frac{7}{x - 3}

= 5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3}

定数を解答に追加する

= 5 ln ∣ x - 2 ∣ - 5 ln ∣ x - 3 ∣ - \frac{7}{x - 3} + C