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

解

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

\frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

\int \frac{x}{x ( x - 1 ) ( x + 4 )} d x = \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣

\int \frac{3}{x ( x - 1 ) ( x + 4 )} d x = 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣)

= \frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣)

簡素化

\frac{1}{5} ln ∣ x - 1 ∣ - \frac{1}{5} ln ∣ x + 4 ∣ + 3 (\frac{1}{20} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x - 1 ∣) : \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣

= \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣

定数を解答に追加する

= \frac{4}{5} ln ∣ x - 1 ∣ - \frac{1}{20} ln ∣ x + 4 ∣ - \frac{3}{4} ln ∣ x ∣ + C