integral of (x^2+7x+3)^4(2x+7)

解

\int (x^{2} + 7 x + 3)^{4} (2 x + 7) d x

\frac{x ^{10}}{5} + 7 x^{9} + 101 x^{8} + 770 x^{7} + 3301 x^{6} + \frac{39277 x ^{5}}{5} + 9903 x^{4} + 6930 x^{3} + 2727 x^{2} + 567 x + C

解答ステップ

\int (x^{2} + 7 x + 3)^{4} (2 x + 7) d x

拡張

(x^{2} + 7 x + 3)^{4} (2 x + 7) : 2 x^{9} + 63 x^{8} + 808 x^{7} + 5390 x^{6} + 19806 x^{5} + 39277 x^{4} + 39612 x^{3} + 20790 x^{2} + 5454 x + 567

= \int 2 x^{9} + 63 x^{8} + 808 x^{7} + 5390 x^{6} + 19806 x^{5} + 39277 x^{4} + 39612 x^{3} + 20790 x^{2} + 5454 x + 567 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 2 x^{9} d x + \int 63 x^{8} d x + \int 808 x^{7} d x + \int 5390 x^{6} d x + \int 19806 x^{5} d x + \int 39277 x^{4} d x + \int 39612 x^{3} d x + \int 20790 x^{2} d x + \int 5454 x d x + \int 567 d x

\int 2 x^{9} d x = \frac{x ^{10}}{5}

\int 63 x^{8} d x = 7 x^{9}

\int 808 x^{7} d x = 101 x^{8}

\int 5390 x^{6} d x = 770 x^{7}

\int 19806 x^{5} d x = 3301 x^{6}

\int 39277 x^{4} d x = \frac{39277 x ^{5}}{5}

\int 39612 x^{3} d x = 9903 x^{4}

\int 20790 x^{2} d x = 6930 x^{3}

\int 5454 x d x = 2727 x^{2}

\int 567 d x = 567 x

= \frac{x ^{10}}{5} + 7 x^{9} + 101 x^{8} + 770 x^{7} + 3301 x^{6} + \frac{39277 x ^{5}}{5} + 9903 x^{4} + 6930 x^{3} + 2727 x^{2} + 567 x

定数を解答に追加する

= \frac{x ^{10}}{5} + 7 x^{9} + 101 x^{8} + 770 x^{7} + 3301 x^{6} + \frac{39277 x ^{5}}{5} + 9903 x^{4} + 6930 x^{3} + 2727 x^{2} + 567 x + C