integral of (x^3+x+1)^4(3x^2+1)

解

\int (x^{3} + x + 1)^{4} (3 x^{2} + 1) d x

\frac{x ^{15}}{5} + x^{13} + x^{12} + 2 x^{11} + 4 x^{10} + 4 x^{9} + 6 x^{8} + 7 x^{7} + 6 x^{6} + \frac{31 x ^{5}}{5} + 5 x^{4} + 3 x^{3} + 2 x^{2} + x + C

解答ステップ

\int (x^{3} + x + 1)^{4} (3 x^{2} + 1) d x

拡張

(x^{3} + x + 1)^{4} (3 x^{2} + 1) : 3 x^{14} + 13 x^{12} + 12 x^{11} + 22 x^{10} + 40 x^{9} + 36 x^{8} + 48 x^{7} + 49 x^{6} + 36 x^{5} + 31 x^{4} + 20 x^{3} + 9 x^{2} + 4 x + 1

= \int 3 x^{14} + 13 x^{12} + 12 x^{11} + 22 x^{10} + 40 x^{9} + 36 x^{8} + 48 x^{7} + 49 x^{6} + 36 x^{5} + 31 x^{4} + 20 x^{3} + 9 x^{2} + 4 x + 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 3 x^{14} d x + \int 13 x^{12} d x + \int 12 x^{11} d x + \int 22 x^{10} d x + \int 40 x^{9} d x + \int 36 x^{8} d x + \int 48 x^{7} d x + \int 49 x^{6} d x + \int 36 x^{5} d x + \int 31 x^{4} d x + \int 20 x^{3} d x + \int 9 x^{2} d x + \int 4 x d x + \int 1 d x

\int 3 x^{14} d x = \frac{x ^{15}}{5}

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

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

\int 22 x^{10} d x = 2 x^{11}

\int 40 x^{9} d x = 4 x^{10}

\int 36 x^{8} d x = 4 x^{9}

\int 48 x^{7} d x = 6 x^{8}

\int 49 x^{6} d x = 7 x^{7}

\int 36 x^{5} d x = 6 x^{6}

\int 31 x^{4} d x = \frac{31 x ^{5}}{5}

\int 20 x^{3} d x = 5 x^{4}

\int 9 x^{2} d x = 3 x^{3}

\int 4 x d x = 2 x^{2}

\int 1 d x = x

= \frac{x ^{15}}{5} + x^{13} + x^{12} + 2 x^{11} + 4 x^{10} + 4 x^{9} + 6 x^{8} + 7 x^{7} + 6 x^{6} + \frac{31 x ^{5}}{5} + 5 x^{4} + 3 x^{3} + 2 x^{2} + x

定数を解答に追加する

= \frac{x ^{15}}{5} + x^{13} + x^{12} + 2 x^{11} + 4 x^{10} + 4 x^{9} + 6 x^{8} + 7 x^{7} + 6 x^{6} + \frac{31 x ^{5}}{5} + 5 x^{4} + 3 x^{3} + 2 x^{2} + x + C