integral of (x^2+4x-5)/(x^3)

解

\int \frac{x ^{2} + 4 x - 5}{x ^{3}} d x

ln ∣ x ∣ - \frac{4}{x} + \frac{5}{2 x ^{2}} + C

解答ステップ

\int \frac{x ^{2} + 4 x - 5}{x ^{3}} d x

拡張

\frac{x ^{2} + 4 x - 5}{x ^{3}} : \frac{1}{x} + \frac{4}{x ^{2}} - \frac{5}{x ^{3}}

= \int \frac{1}{x} + \frac{4}{x ^{2}} - \frac{5}{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{4}{x ^{2}} d x - \int \frac{5}{x ^{3}} d x

\int \frac{1}{x} d x = ln ∣ x ∣

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

\int \frac{5}{x ^{3}} d x = - \frac{5}{2 x ^{2}}

= ln ∣ x ∣ - \frac{4}{x} - (- \frac{5}{2 x ^{2}})

簡素化

= ln ∣ x ∣ - \frac{4}{x} + \frac{5}{2 x ^{2}}

定数を解答に追加する

= ln ∣ x ∣ - \frac{4}{x} + \frac{5}{2 x ^{2}} + C