integral of (3x^2-x+5)/(x^4)

解

\int \frac{3 x ^{2} - x + 5}{x ^{4}} d x

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

解答ステップ

\int \frac{3 x ^{2} - x + 5}{x ^{4}} d x

拡張

\frac{3 x ^{2} - x + 5}{x ^{4}} : \frac{3}{x ^{2}} - \frac{1}{x ^{3}} + \frac{5}{x ^{4}}

= \int \frac{3}{x ^{2}} - \frac{1}{x ^{3}} + \frac{5}{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{3}{x ^{2}} d x - \int \frac{1}{x ^{3}} d x + \int \frac{5}{x ^{4}} d x

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

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

\int \frac{5}{x ^{4}} d x = - \frac{5}{3 x ^{3}}

= - \frac{3}{x} - (- \frac{1}{2 x ^{2}}) - \frac{5}{3 x ^{3}}

簡素化

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

定数を解答に追加する

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