integral of (x^2-1)/(x^3-5x^2+x-5)

解

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

\frac{12}{13} ln ∣ x - 5 ∣ + \frac{1}{13} (\frac{1}{2} ln x^{2} + 1 + 5 arctan (x)) + C

解答ステップ

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

以下の部分分数を得る:

\frac{x ^{2} - 1}{x ^{3} - 5 x ^{2} + x - 5} : \frac{12}{13 ( x - 5 )} + \frac{x + 5}{13 ( x ^{2} + 1 )}

= \int \frac{12}{13 ( x - 5 )} + \frac{x + 5}{13 ( x ^{2} + 1 )} 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{12}{13 ( x - 5 )} d x + \int \frac{x + 5}{13 ( x ^{2} + 1 )} d x

\int \frac{12}{13 ( x - 5 )} d x = \frac{12}{13} ln ∣ x - 5 ∣

\int \frac{x + 5}{13 ( x ^{2} + 1 )} d x = \frac{1}{13} (\frac{1}{2} ln x^{2} + 1 + 5 arctan (x))

= \frac{12}{13} ln ∣ x - 5 ∣ + \frac{1}{13} (\frac{1}{2} ln x^{2} + 1 + 5 arctan (x))

定数を解答に追加する

= \frac{12}{13} ln ∣ x - 5 ∣ + \frac{1}{13} (\frac{1}{2} ln x^{2} + 1 + 5 arctan (x)) + C