integral of 2/(x^3-5x^2+3x+9)

解

\int \frac{2}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) + C

解答ステップ

\int \frac{2}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{x ^{3} - 5 x ^{2} + 3 x + 9} d x

以下の部分分数を得る:

\frac{1}{x ^{3} - 5 x ^{2} + 3 x + 9} : \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 3 )} + \frac{1}{4 ( x - 3 ) ^{2}}

= 2 \cdot \int \frac{1}{16 ( x + 1 )} - \frac{1}{16 ( x - 3 )} + \frac{1}{4 ( 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

= 2 (\int \frac{1}{16 ( x + 1 )} d x - \int \frac{1}{16 ( x - 3 )} d x + \int \frac{1}{4 ( x - 3 ) ^{2}} d x)

\int \frac{1}{16 ( x + 1 )} d x = \frac{1}{16} ln ∣ x + 1 ∣

\int \frac{1}{16 ( x - 3 )} d x = \frac{1}{16} ln ∣ x - 3 ∣

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

= 2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )})

定数を解答に追加する

= 2 (\frac{1}{16} ln ∣ x + 1 ∣ - \frac{1}{16} ln ∣ x - 3 ∣ - \frac{1}{4 ( x - 3 )}) + C