integral of (x^5+2x^2-1)/(x^4)

解

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

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

解答ステップ

\int \frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} d x

拡張

\frac{x ^{5} + 2 x ^{2} - 1}{x ^{4}} : x + \frac{2}{x ^{2}} - \frac{1}{x ^{4}}

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

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

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

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

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

簡素化

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

定数を解答に追加する

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