integral of (3+5x-6x^2-7x^3)/(2x^2)

解

\int \frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{2 x ^{2}} d x

\frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2}) + C

解答ステップ

\int \frac{3 + 5 x - 6 x ^{2} - 7 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{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{x ^{2}} d x

拡張

\frac{3 + 5 x - 6 x ^{2} - 7 x ^{3}}{x ^{2}} : \frac{3}{x ^{2}} + \frac{5}{x} - 6 - 7 x

= \frac{1}{2} \cdot \int \frac{3}{x ^{2}} + \frac{5}{x} - 6 - 7 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}{2} (\int \frac{3}{x ^{2}} d x + \int \frac{5}{x} d x - \int 6 d x - \int 7 x d x)

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

\int \frac{5}{x} d x = 5 ln ∣ x ∣

\int 6 d x = 6 x

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

= \frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2})

定数を解答に追加する

= \frac{1}{2} (- \frac{3}{x} + 5 ln ∣ x ∣ - 6 x - \frac{7 x ^{2}}{2}) + C