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

解

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

\frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{x + 1}{x ^{2} - x - 6} : \frac{1}{5 ( x + 2 )} + \frac{4}{5 ( x - 3 )}

= \int \frac{1}{5 ( x + 2 )} + \frac{4}{5 ( 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}{5 ( x + 2 )} d x + \int \frac{4}{5 ( x - 3 )} d x

\int \frac{1}{5 ( x + 2 )} d x = \frac{1}{5} ln ∣ x + 2 ∣

\int \frac{4}{5 ( x - 3 )} d x = \frac{4}{5} ln ∣ x - 3 ∣

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣

定数を解答に追加する

= \frac{1}{5} ln ∣ x + 2 ∣ + \frac{4}{5} ln ∣ x - 3 ∣ + C