integral of (x+1)/(x^3+5x^2+6x)

解

\int \frac{x + 1}{x ^{3} + 5 x ^{2} + 6 x} d x

\frac{1}{6} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ - \frac{2}{3} ln ∣ x + 3 ∣ + C

解答ステップ

\int \frac{x + 1}{x ^{3} + 5 x ^{2} + 6 x} d x

以下の部分分数を得る:

\frac{x + 1}{x ^{3} + 5 x ^{2} + 6 x} : \frac{1}{6 x} + \frac{1}{2 ( x + 2 )} - \frac{2}{3 ( x + 3 )}

= \int \frac{1}{6 x} + \frac{1}{2 ( x + 2 )} - \frac{2}{3 ( 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{1}{6 x} d x + \int \frac{1}{2 ( x + 2 )} d x - \int \frac{2}{3 ( x + 3 )} d x

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

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

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

= \frac{1}{6} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ - \frac{2}{3} ln ∣ x + 3 ∣

定数を解答に追加する

= \frac{1}{6} ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 2 ∣ - \frac{2}{3} ln ∣ x + 3 ∣ + C