integral of 8x^3-5x^{1/3}+4

解

\int 8 x^{3} - 5 x^{\frac{1}{3}} + 4 d x

2 x^{4} - \frac{15}{4} x^{\frac{4}{3}} + 4 x + C

解答ステップ

\int 8 x^{3} - 5 x^{\frac{1}{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 8 x^{3} d x - \int 5 x^{\frac{1}{3}} d x + \int 4 d x

\int 8 x^{3} d x = 2 x^{4}

\int 5 x^{\frac{1}{3}} d x = \frac{15}{4} x^{\frac{4}{3}}

\int 4 d x = 4 x

= 2 x^{4} - \frac{15}{4} x^{\frac{4}{3}} + 4 x

定数を解答に追加する

= 2 x^{4} - \frac{15}{4} x^{\frac{4}{3}} + 4 x + C