integral of 4/(x^2)+2-1/(8x^3)

解

\int \frac{4}{x ^{2}} + 2 - \frac{1}{8 x ^{3}} d x

- \frac{4}{x} + 2 x + \frac{1}{16 x ^{2}} + C

解答ステップ

\int \frac{4}{x ^{2}} + 2 - \frac{1}{8 x ^{3}} 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{4}{x ^{2}} d x + \int 2 d x - \int \frac{1}{8 x ^{3}} d x

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

\int 2 d x = 2 x

\int \frac{1}{8 x ^{3}} d x = - \frac{1}{16 x ^{2}}

= - \frac{4}{x} + 2 x - (- \frac{1}{16 x ^{2}})

簡素化

= - \frac{4}{x} + 2 x + \frac{1}{16 x ^{2}}

定数を解答に追加する

= - \frac{4}{x} + 2 x + \frac{1}{16 x ^{2}} + C