integral of 30(x+5)(x+7)^4

解

\int 30 (x + 5) (x + 7)^{4} d x

30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x) + C

解答ステップ

\int 30 (x + 5) (x + 7)^{4} d x

定数を除く:

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

= 30 \cdot \int (x + 5) (x + 7)^{4} d x

拡張

(x + 5) (x + 7)^{4} : x^{5} + 33 x^{4} + 434 x^{3} + 2842 x^{2} + 9261 x + 12005

= 30 \cdot \int x^{5} + 33 x^{4} + 434 x^{3} + 2842 x^{2} + 9261 x + 12005 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

= 30 (\int x^{5} d x + \int 33 x^{4} d x + \int 434 x^{3} d x + \int 2842 x^{2} d x + \int 9261 x d x + \int 12005 d x)

\int x^{5} d x = \frac{x ^{6}}{6}

\int 33 x^{4} d x = \frac{33 x ^{5}}{5}

\int 434 x^{3} d x = \frac{217 x ^{4}}{2}

\int 2842 x^{2} d x = \frac{2842 x ^{3}}{3}

\int 9261 x d x = \frac{9261 x ^{2}}{2}

\int 12005 d x = 12005 x

= 30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x)

定数を解答に追加する

= 30 (\frac{x ^{6}}{6} + \frac{33 x ^{5}}{5} + \frac{217 x ^{4}}{2} + \frac{2842 x ^{3}}{3} + \frac{9261 x ^{2}}{2} + 12005 x) + C