integral of (e-x-e^2)(ex+e-5)

解

\int (e - x - e^{2}) (e x + e - 5) d x

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

解答ステップ

\int (e - x - e^{2}) (e x + e - 5) d x

拡張

(e - x - e^{2}) (e x + e - 5) : - e x^{2} - e^{3} x + e^{2} x - e x + 5 x - e^{3} + 6 e^{2} - 5 e

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

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

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

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

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

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

\int e^{3} d x = e^{3} x

\int 6 e^{2} d x = 6 e^{2} x

\int 5 e d x = 5 e x

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

定数を解答に追加する

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