integral of (5x^2+28x+29)/((x+1)(x+2)^2)

解

\int \frac{5 x ^{2} + 28 x + 29}{( x + 1 ) ( x + 2 ) ^{2}} d x

6 ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ - \frac{7}{x + 2} + C

解答ステップ

\int \frac{5 x ^{2} + 28 x + 29}{( x + 1 ) ( x + 2 ) ^{2}} d x

以下の部分分数を得る:

\frac{5 x ^{2} + 28 x + 29}{( x + 1 ) ( x + 2 ) ^{2}} : \frac{6}{x + 1} - \frac{1}{x + 2} + \frac{7}{( x + 2 ) ^{2}}

= \int \frac{6}{x + 1} - \frac{1}{x + 2} + \frac{7}{( x + 2 ) ^{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{6}{x + 1} d x - \int \frac{1}{x + 2} d x + \int \frac{7}{( x + 2 ) ^{2}} d x

\int \frac{6}{x + 1} d x = 6 ln ∣ x + 1 ∣

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

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

= 6 ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ - \frac{7}{x + 2}

定数を解答に追加する

= 6 ln ∣ x + 1 ∣ - ln ∣ x + 2 ∣ - \frac{7}{x + 2} + C