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

解

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

\frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C

解答ステップ

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

拡張

x^{3} (x^{4} + 3)^{2} : x^{11} + 6 x^{7} + 9 x^{3}

= \int x^{11} + 6 x^{7} + 9 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 x^{11} d x + \int 6 x^{7} d x + \int 9 x^{3} d x

\int x^{11} d x = \frac{x ^{12}}{12}

\int 6 x^{7} d x = \frac{3 x ^{8}}{4}

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

= \frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4}

定数を解答に追加する

= \frac{x ^{12}}{12} + \frac{3 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C