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

解

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

2 ln ∣ x ∣ + \frac{1}{x} + C

解答ステップ

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

拡張

\frac{2 x - 1}{x ^{2}} : \frac{2}{x} - \frac{1}{x ^{2}}

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 ln ∣ x ∣ - (- \frac{1}{x})

簡素化

= 2 ln ∣ x ∣ + \frac{1}{x}

定数を解答に追加する

= 2 ln ∣ x ∣ + \frac{1}{x} + C