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

解

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

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

解答ステップ

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

拡張

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

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

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

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

簡素化

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

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

定数を解答に追加する

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