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

解

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

3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) + C

解答ステップ

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

以下の部分分数を得る:

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

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

\int \frac{1}{4 ( x - 3 )} d x = \frac{1}{4} ln ∣ x - 3 ∣

= 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

定数を解答に追加する

= 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) + C