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

解

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

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

解答ステップ

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

拡張

\frac{( 3 - x ) ( 5 + x )}{x ^{4}} : \frac{15}{x ^{4}} - \frac{2}{x ^{3}} - \frac{1}{x ^{2}}

= \int \frac{15}{x ^{4}} - \frac{2}{x ^{3}} - \frac{1}{x ^{2}} 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{15}{x ^{4}} d x - \int \frac{2}{x ^{3}} d x - \int \frac{1}{x ^{2}} d x

\int \frac{15}{x ^{4}} d x = - \frac{5}{x ^{3}}

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

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

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

簡素化

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

定数を解答に追加する

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