integral of 4x^2(2x^3-5)^6

解

\int 4 x^{2} (2 x^{3} - 5)^{6} d x

4 (\frac{64 x ^{21}}{21} - \frac{160 x ^{18}}{3} + 400 x^{15} - \frac{5000 x ^{12}}{3} + \frac{12500 x ^{9}}{3} - 6250 x^{6} + \frac{15625 x ^{3}}{3}) + C

解答ステップ

\int 4 x^{2} (2 x^{3} - 5)^{6} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x^{2} (2 x^{3} - 5)^{6} d x

拡張

x^{2} (2 x^{3} - 5)^{6} : 64 x^{20} - 960 x^{17} + 6000 x^{14} - 20000 x^{11} + 37500 x^{8} - 37500 x^{5} + 15625 x^{2}

= 4 \cdot \int 64 x^{20} - 960 x^{17} + 6000 x^{14} - 20000 x^{11} + 37500 x^{8} - 37500 x^{5} + 15625 x^{2} 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

= 4 (\int 64 x^{20} d x - \int 960 x^{17} d x + \int 6000 x^{14} d x - \int 20000 x^{11} d x + \int 37500 x^{8} d x - \int 37500 x^{5} d x + \int 15625 x^{2} d x)

\int 64 x^{20} d x = \frac{64 x ^{21}}{21}

\int 960 x^{17} d x = \frac{160 x ^{18}}{3}

\int 6000 x^{14} d x = 400 x^{15}

\int 20000 x^{11} d x = \frac{5000 x ^{12}}{3}

\int 37500 x^{8} d x = \frac{12500 x ^{9}}{3}

\int 37500 x^{5} d x = 6250 x^{6}

\int 15625 x^{2} d x = \frac{15625 x ^{3}}{3}

= 4 (\frac{64 x ^{21}}{21} - \frac{160 x ^{18}}{3} + 400 x^{15} - \frac{5000 x ^{12}}{3} + \frac{12500 x ^{9}}{3} - 6250 x^{6} + \frac{15625 x ^{3}}{3})

定数を解答に追加する

= 4 (\frac{64 x ^{21}}{21} - \frac{160 x ^{18}}{3} + 400 x^{15} - \frac{5000 x ^{12}}{3} + \frac{12500 x ^{9}}{3} - 6250 x^{6} + \frac{15625 x ^{3}}{3}) + C