integral of (x^5+1/(x^3)+4)

解

\int (x^{5} + \frac{1}{x ^{3}} + 4) d x

\frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x + C

解答ステップ

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

\int x^{5} d x = \frac{x ^{6}}{6}

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

\int 4 d x = 4 x

= \frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x

定数を解答に追加する

= \frac{x ^{6}}{6} - \frac{1}{2 x ^{2}} + 4 x + C