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

解

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

- \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1)) + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{2} - 2 x - 3}{( x - 1 ) ( x ^{2} + 2 x + 2 )} : - \frac{4}{5 ( x - 1 )} + \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )}

= \int - \frac{4}{5 ( x - 1 )} + \frac{9 x + 7}{5 ( x ^{2} + 2 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{4}{5 ( x - 1 )} d x + \int \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )} d x

\int \frac{4}{5 ( x - 1 )} d x = \frac{4}{5} ln ∣ x - 1 ∣

\int \frac{9 x + 7}{5 ( x ^{2} + 2 x + 2 )} d x = \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1))

= - \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1))

定数を解答に追加する

= - \frac{4}{5} ln ∣ x - 1 ∣ + \frac{1}{10} (9 ln x^{2} + 2 x + 2 - 4 arctan (x + 1)) + C