integral of (6x^3+1)/(x^2)

解

\int \frac{6 x ^{3} + 1}{x ^{2}} d x

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

解答ステップ

\int \frac{6 x ^{3} + 1}{x ^{2}} d x

拡張

\frac{6 x ^{3} + 1}{x ^{2}} : 6 x + \frac{1}{x ^{2}}

= \int 6 x + \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 6 x d x + \int \frac{1}{x ^{2}} d x

\int 6 x d x = 3 x^{2}

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

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

定数を解答に追加する

= 3 x^{2} - \frac{1}{x} + C