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

解

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

- \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣ + C

解答ステップ

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

拡張

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

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

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

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= - \frac{1}{x ^{2}} - (- \frac{3}{x}) + 4 ln ∣ x ∣

簡素化

= - \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣

定数を解答に追加する

= - \frac{1}{x ^{2}} + \frac{3}{x} + 4 ln ∣ x ∣ + C