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

解

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

x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{1}{21} (2 x - 5) + C

解答ステップ

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

拡張

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

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

\int \frac{x}{x ^{2} - 5 x + 1} d x = x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{4 x ^{2} - 20 x + 4}{21} + \frac{2}{21} x - \frac{5}{21}

\int \frac{3}{x ^{2} - 5 x + 1} d x = 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21}

= x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{4 x ^{2} - 20 x + 4}{21} + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21}

簡素化

x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{4 x ^{2} - 20 x + 4}{21} + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21} : x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{1}{21} (2 x - 5)

= x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{1}{21} (2 x - 5)

定数を解答に追加する

= x^{2} - 5 x + 1 + \frac{5}{2} ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{2}{21} x - \frac{5}{21} - 3 ln \frac{1}{21} (4 x^{2} - 20 x + 4) + \frac{1}{21} (2 x - 5) + C