integral of (x+1)/(x^3-6x^2+9x)

解

\int \frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} d x

\frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )} + C

解答ステップ

\int \frac{x + 1}{x ^{3} - 6 x ^{2} + 9 x} d x

以下の部分分数を得る:

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

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

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

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

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

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )}

定数を解答に追加する

= \frac{1}{9} ln ∣ x ∣ - \frac{1}{9} ln ∣ x - 3 ∣ - \frac{4}{3 ( x - 3 )} + C