integral of (5x^2+2)^610

解

\int (5 x^{2} + 2)^{6} 10 d x

10 (\frac{15625 x ^{13}}{13} + \frac{37500 x ^{11}}{11} + \frac{12500 x ^{9}}{3} + \frac{20000 x ^{7}}{7} + 1200 x^{5} + 320 x^{3} + 64 x) + C

解答ステップ

\int (5 x^{2} + 2)^{6} \cdot 10 d x

定数を除く:

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

= 10 \cdot \int (5 x^{2} + 2)^{6} d x

拡張

(5 x^{2} + 2)^{6} : 15625 x^{12} + 37500 x^{10} + 37500 x^{8} + 20000 x^{6} + 6000 x^{4} + 960 x^{2} + 64

= 10 \cdot \int 15625 x^{12} + 37500 x^{10} + 37500 x^{8} + 20000 x^{6} + 6000 x^{4} + 960 x^{2} + 64 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

= 10 (\int 15625 x^{12} d x + \int 37500 x^{10} d x + \int 37500 x^{8} d x + \int 20000 x^{6} d x + \int 6000 x^{4} d x + \int 960 x^{2} d x + \int 64 d x)

\int 15625 x^{12} d x = \frac{15625 x ^{13}}{13}

\int 37500 x^{10} d x = \frac{37500 x ^{11}}{11}

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

\int 20000 x^{6} d x = \frac{20000 x ^{7}}{7}

\int 6000 x^{4} d x = 1200 x^{5}

\int 960 x^{2} d x = 320 x^{3}

\int 64 d x = 64 x

= 10 (\frac{15625 x ^{13}}{13} + \frac{37500 x ^{11}}{11} + \frac{12500 x ^{9}}{3} + \frac{20000 x ^{7}}{7} + 1200 x^{5} + 320 x^{3} + 64 x)

定数を解答に追加する

= 10 (\frac{15625 x ^{13}}{13} + \frac{37500 x ^{11}}{11} + \frac{12500 x ^{9}}{3} + \frac{20000 x ^{7}}{7} + 1200 x^{5} + 320 x^{3} + 64 x) + C