integral of (x^3+1)^{10}

解

\int (x^{3} + 1)^{10} d x

\frac{x ^{31}}{31} + \frac{5 x ^{28}}{14} + \frac{9 x ^{25}}{5} + \frac{60 x ^{22}}{11} + \frac{210 x ^{19}}{19} + \frac{63 x ^{16}}{4} + \frac{210 x ^{13}}{13} + 12 x^{10} + \frac{45 x ^{7}}{7} + \frac{5 x ^{4}}{2} + x + C

解答ステップ

\int (x^{3} + 1)^{10} d x

拡張

(x^{3} + 1)^{10} : x^{30} + 10 x^{27} + 45 x^{24} + 120 x^{21} + 210 x^{18} + 252 x^{15} + 210 x^{12} + 120 x^{9} + 45 x^{6} + 10 x^{3} + 1

= \int x^{30} + 10 x^{27} + 45 x^{24} + 120 x^{21} + 210 x^{18} + 252 x^{15} + 210 x^{12} + 120 x^{9} + 45 x^{6} + 10 x^{3} + 1 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^{30} d x + \int 10 x^{27} d x + \int 45 x^{24} d x + \int 120 x^{21} d x + \int 210 x^{18} d x + \int 252 x^{15} d x + \int 210 x^{12} d x + \int 120 x^{9} d x + \int 45 x^{6} d x + \int 10 x^{3} d x + \int 1 d x

\int x^{30} d x = \frac{x ^{31}}{31}

\int 10 x^{27} d x = \frac{5 x ^{28}}{14}

\int 45 x^{24} d x = \frac{9 x ^{25}}{5}

\int 120 x^{21} d x = \frac{60 x ^{22}}{11}

\int 210 x^{18} d x = \frac{210 x ^{19}}{19}

\int 252 x^{15} d x = \frac{63 x ^{16}}{4}

\int 210 x^{12} d x = \frac{210 x ^{13}}{13}

\int 120 x^{9} d x = 12 x^{10}

\int 45 x^{6} d x = \frac{45 x ^{7}}{7}

\int 10 x^{3} d x = \frac{5 x ^{4}}{2}

\int 1 d x = x

= \frac{x ^{31}}{31} + \frac{5 x ^{28}}{14} + \frac{9 x ^{25}}{5} + \frac{60 x ^{22}}{11} + \frac{210 x ^{19}}{19} + \frac{63 x ^{16}}{4} + \frac{210 x ^{13}}{13} + 12 x^{10} + \frac{45 x ^{7}}{7} + \frac{5 x ^{4}}{2} + x

定数を解答に追加する

= \frac{x ^{31}}{31} + \frac{5 x ^{28}}{14} + \frac{9 x ^{25}}{5} + \frac{60 x ^{22}}{11} + \frac{210 x ^{19}}{19} + \frac{63 x ^{16}}{4} + \frac{210 x ^{13}}{13} + 12 x^{10} + \frac{45 x ^{7}}{7} + \frac{5 x ^{4}}{2} + x + C