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

解

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

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

解答ステップ

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

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

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

\int 4 d x = 4 x

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

定数を解答に追加する

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