integral of (x^3-3x+1)/(y+sqrt(y))

解

\int \frac{x ^{3} - 3 x + 1}{y + y} d x

\frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2}) + C

解答ステップ

\int \frac{x ^{3} - 3 x + 1}{y + y} d x

定数を除く:

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

= \frac{1}{y + y} \cdot \int 1 + x^{3} - 3 x d x

簡素化

\frac{1}{y + y} : \frac{1 - y}{y ( 1 - y )}

= \frac{1 - y}{y ( 1 - y )} \cdot \int 1 + x^{3} - 3 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

= \frac{1 - y}{y ( 1 - y )} (\int 1 d x + \int x^{3} d x - \int 3 x d x)

\int 1 d x = x

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

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

= \frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2})

定数を解答に追加する

= \frac{1 - y}{y ( 1 - y )} (x + \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2}) + C