integral of (1-2x^2-4/(x^{-3)})x^{-3}

解

\int (1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} d x

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

解答ステップ

\int (1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} d x

簡素化

(1 - 2 x^{2} - \frac{4}{x ^{- 3}}) x^{- 3} : \frac{1 - 2 x ^{2} - 4 x ^{3}}{x ^{3}}

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

拡張

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

\int 4 d x = 4 x

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

定数を解答に追加する

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