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

解

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

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

解答ステップ

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

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

\int 8 x d x = 4 x^{2}

\int 4 d x = 4 x

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

定数を解答に追加する

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