integral of 6x^8(x^9+3)^5

解

\int 6 x^{8} (x^{9} + 3)^{5} d x

6 (\frac{x ^{54}}{54} + \frac{x ^{45}}{3} + \frac{5 x ^{36}}{2} + 10 x^{27} + \frac{45 x ^{18}}{2} + 27 x^{9}) + C

解答ステップ

\int 6 x^{8} (x^{9} + 3)^{5} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x^{8} (x^{9} + 3)^{5} d x

拡張

x^{8} (x^{9} + 3)^{5} : x^{53} + 15 x^{44} + 90 x^{35} + 270 x^{26} + 405 x^{17} + 243 x^{8}

= 6 \cdot \int x^{53} + 15 x^{44} + 90 x^{35} + 270 x^{26} + 405 x^{17} + 243 x^{8} 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

= 6 (\int x^{53} d x + \int 15 x^{44} d x + \int 90 x^{35} d x + \int 270 x^{26} d x + \int 405 x^{17} d x + \int 243 x^{8} d x)

\int x^{53} d x = \frac{x ^{54}}{54}

\int 15 x^{44} d x = \frac{x ^{45}}{3}

\int 90 x^{35} d x = \frac{5 x ^{36}}{2}

\int 270 x^{26} d x = 10 x^{27}

\int 405 x^{17} d x = \frac{45 x ^{18}}{2}

\int 243 x^{8} d x = 27 x^{9}

= 6 (\frac{x ^{54}}{54} + \frac{x ^{45}}{3} + \frac{5 x ^{36}}{2} + 10 x^{27} + \frac{45 x ^{18}}{2} + 27 x^{9})

定数を解答に追加する

= 6 (\frac{x ^{54}}{54} + \frac{x ^{45}}{3} + \frac{5 x ^{36}}{2} + 10 x^{27} + \frac{45 x ^{18}}{2} + 27 x^{9}) + C