integral of (u^5-3u^4-u^2+2/7)

解

\int (u^{5} - 3 u^{4} - u^{2} + \frac{2}{7}) d u

\frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C

解答ステップ

\int u^{5} - 3 u^{4} - u^{2} + \frac{2}{7} d u

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 u^{5} d u - \int 3 u^{4} d u - \int u^{2} d u + \int \frac{2}{7} d u

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

\int 3 u^{4} d u = \frac{3 u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

\int \frac{2}{7} d u = \frac{2}{7} u

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u

定数を解答に追加する

= \frac{u ^{6}}{6} - \frac{3 u ^{5}}{5} - \frac{u ^{3}}{3} + \frac{2}{7} u + C