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

解

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

ln ∣ x + 1 ∣ + 2 arctan (\frac{1}{2} (x - 1)) + C

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

= ln ∣ x + 1 ∣ + 2 arctan (\frac{1}{2} (x - 1))

定数を解答に追加する

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