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

解

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

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

解答ステップ

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

\int \frac{- 1}{x} d x = - ln ∣ x ∣

\int ln (2) d x = ln (2) x

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

= - ln ∣ x ∣ - ln (2) x - \frac{1}{x}

定数を解答に追加する

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