integral of 1/(9x^3-2x^2)

解

\int \frac{1}{9 x ^{3} - 2 x ^{2}} d x

- \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣ + C

解答ステップ

\int \frac{1}{9 x ^{3} - 2 x ^{2}} d x

以下の部分分数を得る:

\frac{1}{9 x ^{3} - 2 x ^{2}} : - \frac{9}{4 x} - \frac{1}{2 x ^{2}} + \frac{81}{4 ( 9 x - 2 )}

= \int - \frac{9}{4 x} - \frac{1}{2 x ^{2}} + \frac{81}{4 ( 9 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{9}{4 x} d x - \int \frac{1}{2 x ^{2}} d x + \int \frac{81}{4 ( 9 x - 2 )} d x

\int \frac{9}{4 x} d x = \frac{9}{4} ln ∣ x ∣

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

\int \frac{81}{4 ( 9 x - 2 )} d x = \frac{9}{4} ln ∣ 9 x - 2 ∣

= - \frac{9}{4} ln ∣ x ∣ - (- \frac{1}{2 x}) + \frac{9}{4} ln ∣ 9 x - 2 ∣

簡素化

= - \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣

定数を解答に追加する

= - \frac{9}{4} ln ∣ x ∣ + \frac{1}{2 x} + \frac{9}{4} ln ∣ 9 x - 2 ∣ + C