integral of (x-5)/(x^3-3x^2+7x-5)

解

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

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

解答ステップ

\int \frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

拡張

\frac{x - 5}{x ^{3} - 3 x ^{2} + 7 x - 5} : \frac{x}{x ^{3} - 3 x ^{2} + 7 x - 5} - \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5}

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}{x ^{3} - 3 x ^{2} + 7 x - 5} d x - \int \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x

\int \frac{x}{x ^{3} - 3 x ^{2} + 7 x - 5} d x = \frac{1}{4} ln ∣ x - 1 ∣ + \frac{1}{4} (- \frac{1}{2} ln \frac{1}{4} (x - 1)^{2} + 1 + 2 arctan (\frac{1}{2} (x - 1)))

\int \frac{5}{x ^{3} - 3 x ^{2} + 7 x - 5} d x = 5 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{8} ln x^{2} - 2 x + 5)

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

簡素化

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

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

定数を解答に追加する

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