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

解

\int \frac{2 x + 1}{x ( x - 2 )} d x

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

解答ステップ

\int \frac{2 x + 1}{x ( x - 2 )} d x

拡張

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

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{2 x}{x ( x - 2 )} d x + \int \frac{1}{x ( x - 2 )} d x

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

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

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

簡素化

2 ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x ∣ + \frac{1}{2} ln ∣ x - 2 ∣ : - \frac{1}{2} ln ∣ x ∣ + \frac{5}{2} ln ∣ x - 2 ∣

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

定数を解答に追加する

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