integral of (3x^2+x+1)/(x^2-2x+5)

解

\int \frac{3 x ^{2} + x + 1}{x ^{2} - 2 x + 5} d x

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

解答ステップ

\int \frac{3 x ^{2} + x + 1}{x ^{2} - 2 x + 5} d x

拡張

\frac{3 x ^{2} + x + 1}{x ^{2} - 2 x + 5} : \frac{3 x ^{2}}{x ^{2} - 2 x + 5} + \frac{x}{x ^{2} - 2 x + 5} + \frac{1}{x ^{2} - 2 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{3 x ^{2}}{x ^{2} - 2 x + 5} d x + \int \frac{x}{x ^{2} - 2 x + 5} d x + \int \frac{1}{x ^{2} - 2 x + 5} d x

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

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

\int \frac{1}{x ^{2} - 2 x + 5} d x = \frac{1}{2} arctan (\frac{x - 1}{2})

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

簡素化

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

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

定数を解答に追加する

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