integral of (x^2+4x+1)/((x-1)(x+1)(x+3))

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

\int \frac{1}{4 ( x + 3 )} d x = \frac{1}{4} ln ∣ x + 3 ∣

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

定数を解答に追加する

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