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

解

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

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

解答ステップ

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

拡張

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

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