integral of ((2x^3+10x^2+8x-12)/(4x))

解

\int (\frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{4 x}) d x

\frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣) + C

解答ステップ

\int \frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{4 x} d x

定数を除く:

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

= \frac{1}{4} \cdot \int \frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{x} d x

拡張

\frac{2 x ^{3} + 10 x ^{2} + 8 x - 12}{x} : 2 x^{2} + 10 x + 8 - \frac{12}{x}

= \frac{1}{4} \cdot \int 2 x^{2} + 10 x + 8 - \frac{12}{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}{4} (\int 2 x^{2} d x + \int 10 x d x + \int 8 d x - \int \frac{12}{x} d x)

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

\int 10 x d x = 5 x^{2}

\int 8 d x = 8 x

\int \frac{12}{x} d x = 12 ln ∣ x ∣

= \frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣)

定数を解答に追加する

= \frac{1}{4} (\frac{2 x ^{3}}{3} + 5 x^{2} + 8 x - 12 ln ∣ x ∣) + C