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

解

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

3 (\frac{1}{6} ln ∣ x + 1 ∣ + \frac{2}{15} ln ∣ x - 2 ∣ - \frac{3}{10} ln ∣ x + 3 ∣) + C

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{x ^{3} + 2 x ^{2} - 5 x - 6} d x

以下の部分分数を得る:

\frac{x}{x ^{3} + 2 x ^{2} - 5 x - 6} : \frac{1}{6 ( x + 1 )} + \frac{2}{15 ( x - 2 )} - \frac{3}{10 ( x + 3 )}

= 3 \cdot \int \frac{1}{6 ( x + 1 )} + \frac{2}{15 ( x - 2 )} - \frac{3}{10 ( 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

= 3 (\int \frac{1}{6 ( x + 1 )} d x + \int \frac{2}{15 ( x - 2 )} d x - \int \frac{3}{10 ( x + 3 )} d x)

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

\int \frac{2}{15 ( x - 2 )} d x = \frac{2}{15} ln ∣ x - 2 ∣

\int \frac{3}{10 ( x + 3 )} d x = \frac{3}{10} ln ∣ x + 3 ∣

= 3 (\frac{1}{6} ln ∣ x + 1 ∣ + \frac{2}{15} ln ∣ x - 2 ∣ - \frac{3}{10} ln ∣ x + 3 ∣)

定数を解答に追加する

= 3 (\frac{1}{6} ln ∣ x + 1 ∣ + \frac{2}{15} ln ∣ x - 2 ∣ - \frac{3}{10} ln ∣ x + 3 ∣) + C