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

解

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

\frac{1}{2} x^{2} - 6 x + 11 ln x^{2} + 4 x + 3 + \frac{25}{2} ln ∣ x + 3 ∣ - \frac{25}{2} ln ∣ x + 1 ∣ - 14 + C

解答ステップ

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

拡張

\frac{x ^{3} - 2 x ^{2} + x + 1}{x ^{2} + 4 x + 3} : \frac{x ^{3}}{x ^{2} + 4 x + 3} - \frac{2 x ^{2}}{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{2 x ^{2}}{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{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{13}{2} ln x^{2} + 4 x + 3 + 14 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)

\int \frac{2 x ^{2}}{x ^{2} + 4 x + 3} d x = 2 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣))

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

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

= \frac{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{13}{2} ln x^{2} + 4 x + 3 + 14 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣) - 2 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)) + \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣

簡素化

\frac{1}{2} (x + 2)^{2} - 6 (x + 2) + \frac{13}{2} ln x^{2} + 4 x + 3 + 14 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣) - 2 (x + 2 - 2 ln x^{2} + 4 x + 3 - 5 (\frac{1}{2} ln ∣ x + 3 ∣ - \frac{1}{2} ln ∣ x + 1 ∣)) + \frac{1}{2} ln x^{2} + 4 x + 3 + ln ∣ x + 3 ∣ - ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x + 3 ∣ : \frac{1}{2} x^{2} - 6 x + 11 ln x^{2} + 4 x + 3 + \frac{25}{2} ln ∣ x + 3 ∣ - \frac{25}{2} ln ∣ x + 1 ∣ - 14

= \frac{1}{2} x^{2} - 6 x + 11 ln x^{2} + 4 x + 3 + \frac{25}{2} ln ∣ x + 3 ∣ - \frac{25}{2} ln ∣ x + 1 ∣ - 14

定数を解答に追加する

= \frac{1}{2} x^{2} - 6 x + 11 ln x^{2} + 4 x + 3 + \frac{25}{2} ln ∣ x + 3 ∣ - \frac{25}{2} ln ∣ x + 1 ∣ - 14 + C