integral of (x^3-4x-1)/(x^2)

解

\int \frac{x ^{3} - 4 x - 1}{x ^{2}} d x

\frac{x ^{2}}{2} - 4 ln ∣ x ∣ + \frac{1}{x} + C

解答ステップ

\int \frac{x ^{3} - 4 x - 1}{x ^{2}} d x

拡張

\frac{x ^{3} - 4 x - 1}{x ^{2}} : x - \frac{4}{x} - \frac{1}{x ^{2}}

= \int x - \frac{4}{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 x d x - \int \frac{4}{x} d x - \int \frac{1}{x ^{2}} d x

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

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

= \frac{x ^{2}}{2} - 4 ln ∣ x ∣ - (- \frac{1}{x})

簡素化

= \frac{x ^{2}}{2} - 4 ln ∣ x ∣ + \frac{1}{x}

定数を解答に追加する

= \frac{x ^{2}}{2} - 4 ln ∣ x ∣ + \frac{1}{x} + C