integral of 6x^4+7sqrt(x)+1/(x^3)

解

\int 6 x^{4} + 7 x + \frac{1}{x ^{3}} d x

\frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}} + C

解答ステップ

\int 6 x^{4} + 7 x + \frac{1}{x ^{3}} 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^{4} d x + \int 7 x d x + \int \frac{1}{x ^{3}} d x

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

\int 7 x d x = \frac{14}{3} x^{\frac{3}{2}}

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

= \frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}}

定数を解答に追加する

= \frac{6 x ^{5}}{5} + \frac{14}{3} x^{\frac{3}{2}} - \frac{1}{2 x ^{2}} + C