integral of ((7x^6-8x^4+9x^3)/(2x^2))

解

\int (\frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{2 x ^{2}}) d x

\frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2}) + C

解答ステップ

\int \frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{2 x ^{2}} d x

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{x ^{2}} d x

拡張

\frac{7 x ^{6} - 8 x ^{4} + 9 x ^{3}}{x ^{2}} : \frac{7 x ^{6}}{x ^{2}} - \frac{8 x ^{4}}{x ^{2}} + \frac{9 x ^{3}}{x ^{2}}

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

= \frac{1}{2} (\int \frac{7 x ^{6}}{x ^{2}} d x - \int \frac{8 x ^{4}}{x ^{2}} d x + \int \frac{9 x ^{3}}{x ^{2}} d x)

\int \frac{7 x ^{6}}{x ^{2}} d x = \frac{7 x ^{5}}{5}

\int \frac{8 x ^{4}}{x ^{2}} d x = \frac{8 x ^{3}}{3}

\int \frac{9 x ^{3}}{x ^{2}} d x = \frac{9 x ^{2}}{2}

= \frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2})

定数を解答に追加する

= \frac{1}{2} (\frac{7 x ^{5}}{5} - \frac{8 x ^{3}}{3} + \frac{9 x ^{2}}{2}) + C