integral of (2x^4+6x^3-5x^2+2x-3)

解

\int (2 x^{4} + 6 x^{3} - 5 x^{2} + 2 x - 3) d x

\frac{2 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{5 x ^{3}}{3} + x^{2} - 3 x + C

解答ステップ

\int 2 x^{4} + 6 x^{3} - 5 x^{2} + 2 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 2 x^{4} d x + \int 6 x^{3} d x - \int 5 x^{2} d x + \int 2 x d x - \int 3 d x

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

\int 2 x d x = x^{2}

\int 3 d x = 3 x

= \frac{2 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{5 x ^{3}}{3} + x^{2} - 3 x

定数を解答に追加する

= \frac{2 x ^{5}}{5} + \frac{3 x ^{4}}{2} - \frac{5 x ^{3}}{3} + x^{2} - 3 x + C