integral of (20x^3-40x^2+3)/(x^2-2x)

解

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x

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

解答ステップ

\int \frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} d x

長除法

\frac{20 x ^{3} - 40 x ^{2} + 3}{x ^{2} - 2 x} : 20 x + \frac{3}{x ^{2} - 2 x}

= \int 20 x + \frac{3}{x ^{2} - 2 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

= \int 20 x d x + \int \frac{3}{x ^{2} - 2 x} d x

\int 20 x d x = 10 x^{2}

\int \frac{3}{x ^{2} - 2 x} d x = - 3 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

= 10 x^{2} - 3 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣)

定数を解答に追加する

= 10 x^{2} - 3 (\frac{1}{2} ln ∣ x ∣ - \frac{1}{2} ln ∣ x - 2 ∣) + C