integral of \sqrt[3]{x}-1/(x^2)

解

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

\frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x} + C

解答ステップ

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

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

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

= \frac{3}{4} x^{\frac{4}{3}} - (- \frac{1}{x})

簡素化

= \frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x}

定数を解答に追加する

= \frac{3}{4} x^{\frac{4}{3}} + \frac{1}{x} + C