integral of 1/(6x^3-7x^2+2x)

解

\int \frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} d x

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

解答ステップ

\int \frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} d x

以下の部分分数を得る:

\frac{1}{6 x ^{3} - 7 x ^{2} + 2 x} : \frac{1}{2 x} - \frac{4}{2 x - 1} + \frac{9}{2 ( 3 x - 2 )}

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

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

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

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

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

定数を解答に追加する

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