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

解

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

2 (x + \frac{1}{x} - \frac{1}{x ^{4}}) + C

解答ステップ

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

拡張

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

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

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

\int 1 d x = x

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

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

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

簡素化

= 2 (x + \frac{1}{x} - \frac{1}{x ^{4}})

定数を解答に追加する

= 2 (x + \frac{1}{x} - \frac{1}{x ^{4}}) + C