integral of (2x^2+3x)^3

解

\int (2 x^{2} + 3 x)^{3} d x

\frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4} + C

解答ステップ

\int (2 x^{2} + 3 x)^{3} d x

拡張

(2 x^{2} + 3 x)^{3} : 8 x^{6} + 36 x^{5} + 54 x^{4} + 27 x^{3}

= \int 8 x^{6} + 36 x^{5} + 54 x^{4} + 27 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 8 x^{6} d x + \int 36 x^{5} d x + \int 54 x^{4} d x + \int 27 x^{3} d x

\int 8 x^{6} d x = \frac{8 x ^{7}}{7}

\int 36 x^{5} d x = 6 x^{6}

\int 54 x^{4} d x = \frac{54 x ^{5}}{5}

\int 27 x^{3} d x = \frac{27 x ^{4}}{4}

= \frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4}

定数を解答に追加する

= \frac{8 x ^{7}}{7} + 6 x^{6} + \frac{54 x ^{5}}{5} + \frac{27 x ^{4}}{4} + C