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

解

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

\frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2} + C

解答ステップ

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

簡素化

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

部分積分を適用する

= x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - \int 8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7} d x

\int 8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7} d x = 4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7}

= x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - (4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7})

簡素化

x (8 x + 8 x^{3} + \frac{24 x ^{5}}{5} + \frac{8 x ^{7}}{7}) - (4 x^{2} + 2 x^{4} + \frac{4 x ^{6}}{5} + \frac{x ^{8}}{7}) : \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2}

= \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2}

定数を解答に追加する

= \frac{8 x ^{8}}{7} - \frac{x ^{8}}{7} + \frac{24 x ^{6}}{5} - \frac{4 x ^{6}}{5} + 6 x^{4} + 4 x^{2} + C