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

解

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

- \frac{2}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x - 3 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{2 x ^{3} - 3 x ^{2}} : - \frac{2}{9 x} - \frac{1}{3 x ^{2}} + \frac{4}{9 ( 2 x - 3 )}

= \int - \frac{2}{9 x} - \frac{1}{3 x ^{2}} + \frac{4}{9 ( 2 x - 3 )} 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{2}{9 x} d x - \int \frac{1}{3 x ^{2}} d x + \int \frac{4}{9 ( 2 x - 3 )} d x

\int \frac{2}{9 x} d x = \frac{2}{9} ln ∣ x ∣

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

\int \frac{4}{9 ( 2 x - 3 )} d x = \frac{2}{9} ln ∣ 2 x - 3 ∣

= - \frac{2}{9} ln ∣ x ∣ - (- \frac{1}{3 x}) + \frac{2}{9} ln ∣ 2 x - 3 ∣

簡素化

= - \frac{2}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x - 3 ∣

定数を解答に追加する

= - \frac{2}{9} ln ∣ x ∣ + \frac{1}{3 x} + \frac{2}{9} ln ∣ 2 x - 3 ∣ + C