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

解

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

\frac{8}{7} arctan (\frac{2 x - 1}{7}) + ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} - 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x - 1)) + C

解答ステップ

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

拡張

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

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

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

= \frac{8}{7} arctan (\frac{2 x - 1}{7}) + ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} - 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x - 1))

定数を解答に追加する

= \frac{8}{7} arctan (\frac{2 x - 1}{7}) + ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} - 4 x + 8 + \frac{1}{7} arctan (\frac{1}{7} (2 x - 1)) + C