integral of ((x-1))/(sqrt(x^2-4x+3))

解

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

x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2 + C

解答ステップ

\int \frac{x - 1}{x ^{2} - 4 x + 3} d x

拡張

\frac{x - 1}{x ^{2} - 4 x + 3} : \frac{x}{x ^{2} - 4 x + 3} - \frac{1}{x ^{2} - 4 x + 3}

= \int \frac{x}{x ^{2} - 4 x + 3} - \frac{1}{x ^{2} - 4 x + 3} 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} - 4 x + 3} d x - \int \frac{1}{x ^{2} - 4 x + 3} d x

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

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

= x^{2} - 4 x + 3 + 2 ln x^{2} - 4 x + 3 + x - 2 - ln x^{2} - 4 x + 3 + x - 2

簡素化

= x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2

定数を解答に追加する

= x^{2} - 4 x + 3 + ln x^{2} - 4 x + 3 + x - 2 + C