integral of (x^3+4x^2+8x+5)/(x^2+2x+1)

解

\int \frac{x ^{3} + 4 x ^{2} + 8 x + 5}{x ^{2} + 2 x + 1} d x

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

解答ステップ

\int \frac{x ^{3} + 4 x ^{2} + 8 x + 5}{x ^{2} + 2 x + 1} d x

拡張

\frac{x ^{3} + 4 x ^{2} + 8 x + 5}{x ^{2} + 2 x + 1} : \frac{x ^{3}}{x ^{2} + 2 x + 1} + \frac{4 x ^{2}}{x ^{2} + 2 x + 1} + \frac{8 x}{x ^{2} + 2 x + 1} + \frac{5}{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{x ^{3}}{x ^{2} + 2 x + 1} d x + \int \frac{4 x ^{2}}{x ^{2} + 2 x + 1} d x + \int \frac{8 x}{x ^{2} + 2 x + 1} d x + \int \frac{5}{x ^{2} + 2 x + 1} d x

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

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

\int \frac{5}{x ^{2} + 2 x + 1} d x = - \frac{5}{x + 1}

= \frac{1}{2} (x + 1)^{2} - 3 (x + 1) + 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + 4 (- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣)) + 8 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1}

簡素化

\frac{1}{2} (x + 1)^{2} - 3 (x + 1) + 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} + 4 (- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣)) + 8 (ln ∣ x + 1 ∣ + \frac{1}{x + 1}) - \frac{5}{x + 1} : \frac{1}{2} x^{2} + 6 x + \frac{4}{x + 1} + 3 ln ∣ x + 1 ∣ - \frac{4 x ^{2}}{x + 1} + 5 + \frac{1}{2}

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

定数を解答に追加する

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