integral of (3x^2-6x^2+sqrt(x))/(x^3)

解

\int \frac{3 x ^{2} - 6 x ^{2} + x}{x ^{3}} d x

- 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}} + C

解答ステップ

\int \frac{3 x ^{2} - 6 x ^{2} + x}{x ^{3}} d x

簡素化

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

拡張

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

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

\int \frac{3}{x} d x = 3 ln ∣ x ∣

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

= - 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}}

定数を解答に追加する

= - 3 ln ∣ x ∣ - \frac{2}{3 x ^{\frac{3}{2}}} + C