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

解

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

18 x + \frac{4}{3} (x + 1)^{3} - \frac{x ^{4}}{x + 1} - \frac{3 x ^{2}}{x + 1} - \frac{4}{x + 1} - 10 ln ∣ x + 1 ∣ - 6 (x + 1)^{2} + 18 + C

解答ステップ

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

拡張

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

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

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

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

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

簡素化

- \frac{x ^{4}}{x + 1} + 4 (\frac{1}{3} (x + 1)^{3} - \frac{3}{2} (x + 1)^{2} + 3 (x + 1) - ln ∣ x + 1 ∣) + 3 (- \frac{x ^{2}}{x + 1} + 2 (x + 1 - ln ∣ x + 1 ∣)) - \frac{4}{x + 1} : 18 x + \frac{4}{3} (x + 1)^{3} - \frac{x ^{4}}{x + 1} - \frac{3 x ^{2}}{x + 1} - \frac{4}{x + 1} - 10 ln ∣ x + 1 ∣ - 6 (x + 1)^{2} + 18

= 18 x + \frac{4}{3} (x + 1)^{3} - \frac{x ^{4}}{x + 1} - \frac{3 x ^{2}}{x + 1} - \frac{4}{x + 1} - 10 ln ∣ x + 1 ∣ - 6 (x + 1)^{2} + 18

定数を解答に追加する

= 18 x + \frac{4}{3} (x + 1)^{3} - \frac{x ^{4}}{x + 1} - \frac{3 x ^{2}}{x + 1} - \frac{4}{x + 1} - 10 ln ∣ x + 1 ∣ - 6 (x + 1)^{2} + 18 + C