integral of (3x-4)/((x-5)(x+1))

解

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

\frac{11}{6} ln ∣ x - 5 ∣ + \frac{7}{6} ln ∣ x + 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{3 x - 4}{( x - 5 ) ( x + 1 )} : \frac{11}{6 ( x - 5 )} + \frac{7}{6 ( x + 1 )}

= \int \frac{11}{6 ( x - 5 )} + \frac{7}{6 ( x + 1 )} 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{11}{6 ( x - 5 )} d x + \int \frac{7}{6 ( x + 1 )} d x

\int \frac{11}{6 ( x - 5 )} d x = \frac{11}{6} ln ∣ x - 5 ∣

\int \frac{7}{6 ( x + 1 )} d x = \frac{7}{6} ln ∣ x + 1 ∣

= \frac{11}{6} ln ∣ x - 5 ∣ + \frac{7}{6} ln ∣ x + 1 ∣

定数を解答に追加する

= \frac{11}{6} ln ∣ x - 5 ∣ + \frac{7}{6} ln ∣ x + 1 ∣ + C