integral of (x^2-10x+4)/(x^3)

解

\int \frac{x ^{2} - 10 x + 4}{x ^{3}} d x

ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}} + C

解答ステップ

\int \frac{x ^{2} - 10 x + 4}{x ^{3}} d x

拡張

\frac{x ^{2} - 10 x + 4}{x ^{3}} : \frac{1}{x} - \frac{10}{x ^{2}} + \frac{4}{x ^{3}}

= \int \frac{1}{x} - \frac{10}{x ^{2}} + \frac{4}{x ^{3}} 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 \frac{10}{x ^{2}} d x + \int \frac{4}{x ^{3}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{10}{x ^{2}} d x = - \frac{10}{x}

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

= ln ∣ x ∣ - (- \frac{10}{x}) - \frac{2}{x ^{2}}

簡素化

= ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}}

定数を解答に追加する

= ln ∣ x ∣ + \frac{10}{x} - \frac{2}{x ^{2}} + C