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

解

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

2 x^{2} - 3 x + 1 + C

解答ステップ

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

拡張

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

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

\int \frac{2 x}{x ^{2} - 3 x + 1} d x = 2 x^{2} - 3 x + 1 + 3 ln \frac{1}{5} (4 x^{2} - 12 x + 4) + \frac{2}{5} x - \frac{3}{5}

\int \frac{3}{x ^{2} - 3 x + 1} d x = 3 ln \frac{1}{5} (4 x^{2} - 12 x + 4) + \frac{2}{5} x - \frac{3}{5}

= 2 x^{2} - 3 x + 1 + 3 ln \frac{1}{5} (4 x^{2} - 12 x + 4) + \frac{2}{5} x - \frac{3}{5} - 3 ln \frac{1}{5} (4 x^{2} - 12 x + 4) + \frac{2}{5} x - \frac{3}{5}

簡素化

= 2 x^{2} - 3 x + 1

定数を解答に追加する

= 2 x^{2} - 3 x + 1 + C