integral of (3x^2+4x+2)/(2(y-1))

解

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

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

解答ステップ

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

定数を除く:

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

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

\int 2 d x = 2 x

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

\int 4 x d x = 2 x^{2}

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

簡素化

\frac{1}{( - 1 + y ) \cdot 2} (2 x + x^{3} + 2 x^{2}) : \frac{x}{- 1 + y} + \frac{x ^{3}}{2 ( - 1 + y )} + \frac{x ^{2}}{- 1 + y}

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

定数を解答に追加する

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