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

解

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

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

解答ステップ

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

\int 3 d x = 3 x

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

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

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

簡素化

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

定数を解答に追加する

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