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

解

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

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

解答ステップ

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

拡張

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

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

\int \frac{2 x ^{3}}{x ^{2} + 2 x + 1} d x = x^{2} - 4 x + \frac{2}{x + 1} + 6 ln ∣ x + 1 ∣ - 5

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

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

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

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

簡素化

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

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

定数を解答に追加する

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