integral of x(4x+5)^3

解

\int x (4 x + 5)^{3} d x

\frac{64 x ^{5}}{5} + 60 x^{4} + 100 x^{3} + \frac{125 x ^{2}}{2} + C

解答ステップ

\int x (4 x + 5)^{3} d x

拡張

x (4 x + 5)^{3} : 64 x^{4} + 240 x^{3} + 300 x^{2} + 125 x

= \int 64 x^{4} + 240 x^{3} + 300 x^{2} + 125 x 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 64 x^{4} d x + \int 240 x^{3} d x + \int 300 x^{2} d x + \int 125 x d x

\int 64 x^{4} d x = \frac{64 x ^{5}}{5}

\int 240 x^{3} d x = 60 x^{4}

\int 300 x^{2} d x = 100 x^{3}

\int 125 x d x = \frac{125 x ^{2}}{2}

= \frac{64 x ^{5}}{5} + 60 x^{4} + 100 x^{3} + \frac{125 x ^{2}}{2}

定数を解答に追加する

= \frac{64 x ^{5}}{5} + 60 x^{4} + 100 x^{3} + \frac{125 x ^{2}}{2} + C