integral of (6x^2+13x+6)/((x+2)(x+1)^2)

解

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

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

解答ステップ

\int \frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} d x

以下の部分分数を得る:

\frac{6 x ^{2} + 13 x + 6}{( x + 2 ) ( x + 1 ) ^{2}} : \frac{4}{x + 2} + \frac{2}{x + 1} - \frac{1}{( x + 1 ) ^{2}}

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

\int \frac{4}{x + 2} d x = 4 ln ∣ x + 2 ∣

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

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

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

簡素化

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

定数を解答に追加する

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