integral of (2x^2-3)^5

解

\int (2 x^{2} - 3)^{5} d x

\frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x + C

解答ステップ

\int (2 x^{2} - 3)^{5} d x

拡張

(2 x^{2} - 3)^{5} : 32 x^{10} - 240 x^{8} + 720 x^{6} - 1080 x^{4} + 810 x^{2} - 243

= \int 32 x^{10} - 240 x^{8} + 720 x^{6} - 1080 x^{4} + 810 x^{2} - 243 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 32 x^{10} d x - \int 240 x^{8} d x + \int 720 x^{6} d x - \int 1080 x^{4} d x + \int 810 x^{2} d x - \int 243 d x

\int 32 x^{10} d x = \frac{32 x ^{11}}{11}

\int 240 x^{8} d x = \frac{80 x ^{9}}{3}

\int 720 x^{6} d x = \frac{720 x ^{7}}{7}

\int 1080 x^{4} d x = 216 x^{5}

\int 810 x^{2} d x = 270 x^{3}

\int 243 d x = 243 x

= \frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x

定数を解答に追加する

= \frac{32 x ^{11}}{11} - \frac{80 x ^{9}}{3} + \frac{720 x ^{7}}{7} - 216 x^{5} + 270 x^{3} - 243 x + C