integral of (2x^3+6x^2+7x+4)/(x^2+3x+2)

解

\int \frac{2 x ^{3} + 6 x ^{2} + 7 x + 4}{x ^{2} + 3 x + 2} d x

x^{2} - 3 ln ∣ x + 1 ∣ + 6 ln ∣ x + 2 ∣ - 4 ln ∣ 2 x + 4 ∣ + 4 ln ∣ 2 x + 2 ∣ + C

解答ステップ

\int \frac{2 x ^{3} + 6 x ^{2} + 7 x + 4}{x ^{2} + 3 x + 2} d x

拡張

\frac{2 x ^{3} + 6 x ^{2} + 7 x + 4}{x ^{2} + 3 x + 2} : \frac{2 x ^{3}}{x ^{2} + 3 x + 2} + \frac{6 x ^{2}}{x ^{2} + 3 x + 2} + \frac{7 x}{x ^{2} + 3 x + 2} + \frac{4}{x ^{2} + 3 x + 2}

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{2 x ^{3}}{x ^{2} + 3 x + 2} d x + \int \frac{6 x ^{2}}{x ^{2} + 3 x + 2} d x + \int \frac{7 x}{x ^{2} + 3 x + 2} d x + \int \frac{4}{x ^{2} + 3 x + 2} d x

\int \frac{2 x ^{3}}{x ^{2} + 3 x + 2} d x = x^{2} - 6 x - 2 ln ∣ x + 1 ∣ + 16 ln ∣ x + 2 ∣

\int \frac{6 x ^{2}}{x ^{2} + 3 x + 2} d x = 6 (ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + x)

\int \frac{7 x}{x ^{2} + 3 x + 2} d x = 7 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣)

\int \frac{4}{x ^{2} + 3 x + 2} d x = - 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

= x^{2} - 6 x - 2 ln ∣ x + 1 ∣ + 16 ln ∣ x + 2 ∣ + 6 (ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + x) + 7 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) - 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣)

簡素化

x^{2} - 6 x - 2 ln ∣ x + 1 ∣ + 16 ln ∣ x + 2 ∣ + 6 (ln ∣ x + 1 ∣ - 4 ln ∣ x + 2 ∣ + x) + 7 (- ln ∣ x + 1 ∣ + 2 ln ∣ x + 2 ∣) - 8 (\frac{1}{2} ln ∣ 2 x + 4 ∣ - \frac{1}{2} ln ∣ 2 x + 2 ∣) : x^{2} - 3 ln ∣ x + 1 ∣ + 6 ln ∣ x + 2 ∣ - 4 ln ∣ 2 x + 4 ∣ + 4 ln ∣ 2 x + 2 ∣

= x^{2} - 3 ln ∣ x + 1 ∣ + 6 ln ∣ x + 2 ∣ - 4 ln ∣ 2 x + 4 ∣ + 4 ln ∣ 2 x + 2 ∣

定数を解答に追加する

= x^{2} - 3 ln ∣ x + 1 ∣ + 6 ln ∣ x + 2 ∣ - 4 ln ∣ 2 x + 4 ∣ + 4 ln ∣ 2 x + 2 ∣ + C