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

解

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

\frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣ + C

解答ステップ

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

以下の部分分数を得る:

\frac{1}{( 2 x - 3 ) ( 3 x ^{2} + x )} : \frac{4}{33 ( 2 x - 3 )} - \frac{1}{3 x} + \frac{9}{11 ( 3 x + 1 )}

= \int \frac{4}{33 ( 2 x - 3 )} - \frac{1}{3 x} + \frac{9}{11 ( 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{4}{33 ( 2 x - 3 )} d x - \int \frac{1}{3 x} d x + \int \frac{9}{11 ( 3 x + 1 )} d x

\int \frac{4}{33 ( 2 x - 3 )} d x = \frac{2}{33} ln ∣ 2 x - 3 ∣

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

\int \frac{9}{11 ( 3 x + 1 )} d x = \frac{3}{11} ln ∣ 3 x + 1 ∣

= \frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣

定数を解答に追加する

= \frac{2}{33} ln ∣ 2 x - 3 ∣ - \frac{1}{3} ln ∣ x ∣ + \frac{3}{11} ln ∣ 3 x + 1 ∣ + C