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

解

\int \frac{1}{2 x ^{3} + x ^{2} - x} d x

- ln ∣ x ∣ + \frac{2}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C

解答ステップ

\int \frac{1}{2 x ^{3} + x ^{2} - x} d x

以下の部分分数を得る:

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{4}{3 ( 2 x - 1 )} d x = \frac{2}{3} ln ∣ 2 x - 1 ∣

\int \frac{1}{3 ( x + 1 )} d x = \frac{1}{3} ln ∣ x + 1 ∣

= - ln ∣ x ∣ + \frac{2}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣

定数を解答に追加する

= - ln ∣ x ∣ + \frac{2}{3} ln ∣ 2 x - 1 ∣ + \frac{1}{3} ln ∣ x + 1 ∣ + C