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

解

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

\frac{x ^{2}}{2} + 4 x - 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 + 15 ln ∣ x - 3 ∣ + C

解答ステップ

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

拡張

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

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

\int \frac{x ^{3}}{x ^{2} - 4 x + 3} d x = \frac{x ^{2}}{2} + 4 x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{27}{2} ln ∣ x - 3 ∣

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

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

= \frac{x ^{2}}{2} + 4 x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{27}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 - ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣

簡素化

\frac{x ^{2}}{2} + 4 x - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{27}{2} ln ∣ x - 3 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 - ln ∣ x - 1 ∣ + ln ∣ x - 3 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{2} ln ∣ x - 3 ∣ : \frac{x ^{2}}{2} + 4 x - 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 + 15 ln ∣ x - 3 ∣

= \frac{x ^{2}}{2} + 4 x - 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 + 15 ln ∣ x - 3 ∣

定数を解答に追加する

= \frac{x ^{2}}{2} + 4 x - 2 ln ∣ x - 1 ∣ + \frac{1}{2} ln x^{2} - 4 x + 3 + 15 ln ∣ x - 3 ∣ + C