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

解

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

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

解答ステップ

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

拡張

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

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

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

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

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

定数を解答に追加する

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