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

解

\int \frac{( x ^{2} - 2 x + 5 )}{x ^{2}} d x

x - 2 ln ∣ x ∣ - \frac{5}{x} + C

解答ステップ

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

拡張

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

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

\int 1 d x = x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= x - 2 ln ∣ x ∣ - \frac{5}{x}

定数を解答に追加する

= x - 2 ln ∣ x ∣ - \frac{5}{x} + C