integral of (3/5 x^8-4/3 x^3)^2

解

\int (\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} d x

\frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63} + C

解答ステップ

\int (\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} d x

簡素化

(\frac{3}{5} x^{8} - \frac{4}{3} x^{3})^{2} : \frac{9}{25} x^{16} - \frac{8 x ^{11}}{5} + \frac{16}{9} x^{6}

= \int \frac{9}{25} x^{16} - \frac{8 x ^{11}}{5} + \frac{16}{9} x^{6} 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 \frac{9}{25} x^{16} d x - \int \frac{8 x ^{11}}{5} d x + \int \frac{16}{9} x^{6} d x

\int \frac{9}{25} x^{16} d x = \frac{9 x ^{17}}{425}

\int \frac{8 x ^{11}}{5} d x = \frac{2 x ^{12}}{15}

\int \frac{16}{9} x^{6} d x = \frac{16 x ^{7}}{63}

= \frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63}

定数を解答に追加する

= \frac{9 x ^{17}}{425} - \frac{2 x ^{12}}{15} + \frac{16 x ^{7}}{63} + C