integral of (3x^3+6x)^3(3x^2+2)

解

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

\frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4} + C

解答ステップ

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

拡張

(3 x^{3} + 6 x)^{3} (3 x^{2} + 2) : 81 x^{11} + 540 x^{9} + 1296 x^{7} + 1296 x^{5} + 432 x^{3}

= \int 81 x^{11} + 540 x^{9} + 1296 x^{7} + 1296 x^{5} + 432 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 81 x^{11} d x + \int 540 x^{9} d x + \int 1296 x^{7} d x + \int 1296 x^{5} d x + \int 432 x^{3} d x

\int 81 x^{11} d x = \frac{27 x ^{12}}{4}

\int 540 x^{9} d x = 54 x^{10}

\int 1296 x^{7} d x = 162 x^{8}

\int 1296 x^{5} d x = 216 x^{6}

\int 432 x^{3} d x = 108 x^{4}

= \frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4}

定数を解答に追加する

= \frac{27 x ^{12}}{4} + 54 x^{10} + 162 x^{8} + 216 x^{6} + 108 x^{4} + C