integral of ((4x+3)(2x^2+3x)^3)

解

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

4 x^{8} + 24 x^{7} + 54 x^{6} + 54 x^{5} + \frac{81 x ^{4}}{4} + C

解答ステップ

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

拡張

(4 x + 3) (2 x^{2} + 3 x)^{3} : 32 x^{7} + 168 x^{6} + 324 x^{5} + 270 x^{4} + 81 x^{3}

= \int 32 x^{7} + 168 x^{6} + 324 x^{5} + 270 x^{4} + 81 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 32 x^{7} d x + \int 168 x^{6} d x + \int 324 x^{5} d x + \int 270 x^{4} d x + \int 81 x^{3} d x

\int 32 x^{7} d x = 4 x^{8}

\int 168 x^{6} d x = 24 x^{7}

\int 324 x^{5} d x = 54 x^{6}

\int 270 x^{4} d x = 54 x^{5}

\int 81 x^{3} d x = \frac{81 x ^{4}}{4}

= 4 x^{8} + 24 x^{7} + 54 x^{6} + 54 x^{5} + \frac{81 x ^{4}}{4}

定数を解答に追加する

= 4 x^{8} + 24 x^{7} + 54 x^{6} + 54 x^{5} + \frac{81 x ^{4}}{4} + C