integral of (x^2+8x-3)/(x^3+3x^2)

解

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x ^{2}} d x

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

解答ステップ

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x ^{2}} d x

以下の部分分数を得る:

\frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x ^{2}} : \frac{3}{x} - \frac{1}{x ^{2}} - \frac{2}{x + 3}

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

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

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

簡素化

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

定数を解答に追加する

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