integral of (-2x^2+8x-20)/(x^2-4x+8)

解

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

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

解答ステップ

\int \frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} d x

拡張

\frac{- 2 x ^{2} + 8 x - 20}{x ^{2} - 4 x + 8} : - \frac{2 x ^{2}}{x ^{2} - 4 x + 8} + \frac{8 x}{x ^{2} - 4 x + 8} - \frac{20}{x ^{2} - 4 x + 8}

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{2 x ^{2}}{x ^{2} - 4 x + 8} d x + \int \frac{8 x}{x ^{2} - 4 x + 8} d x - \int \frac{20}{x ^{2} - 4 x + 8} d x

\int \frac{2 x ^{2}}{x ^{2} - 4 x + 8} d x = 2 (2 ln x^{2} - 4 x + 8 + x)

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

\int \frac{20}{x ^{2} - 4 x + 8} d x = 10 arctan (\frac{1}{2} (x - 2))

= - 2 (2 ln x^{2} - 4 x + 8 + x) + 8 (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2))) - 10 arctan (\frac{1}{2} (x - 2))

簡素化

- 2 (2 ln x^{2} - 4 x + 8 + x) + 8 (\frac{1}{2} ln \frac{1}{4} (x - 2)^{2} + 1 + arctan (\frac{1}{2} (x - 2))) - 10 arctan (\frac{1}{2} (x - 2)) : - 4 ln x^{2} - 4 x + 8 - 2 x + 4 ln \frac{1}{4} x^{2} - x + 2 - 2 arctan (\frac{1}{2} x - 1)

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

定数を解答に追加する

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