1-0-x-1-x-2-4x-3-dx-

Question Number 154351 by ZiYangLee last updated on 17/Sep/21

\int_{- 1}^{0} \frac{x - 1}{\sqrt{x^{2} - 4 x + 3}} d x = ?

Answered by ZiYangLee last updated on 17/Sep/21

\int_{- 1}^{0} \frac{x - 1}{\sqrt{x^{2} - 4 x + 3}} d x

= \int_{- 1}^{0} \frac{(x - 2) + 1}{\sqrt{x^{2} - 4 x + 3}} d x

= \int_{- 1}^{0} \frac{\frac{1}{2} (2 x - 4) + 1}{\sqrt{x^{2} - 4 x + 3}} d x

= \int_{- 1}^{0} \frac{(2 x - 4) + 2}{2 \sqrt{x^{2} - 4 x + 3}} d x

= \int_{- 1}^{0} \frac{2 x - 4}{2 \sqrt{x^{2} - 4 x + 3}} d x + \int_{- 1}^{0} \frac{1}{\sqrt{x^{2} - 4 x + 3}} d x

l e t u = x^{2} - 4 x + 3

d u = (2 x - 4) d x

= \int_{8}^{3} \frac{1}{2 \sqrt{u}} d u + \int_{- 1}^{0} {(x^{2} - 4 x + 3)}^{- \frac{1}{2}} d x

= - \int_{3}^{8} \frac{1}{2 \sqrt{u}} d u + {[\frac{\sqrt{x^{2} - 4 x + 3}}{2 (2 x - 4)}]}_{- 1}^{0}

= - \frac{1}{2} {[2 \sqrt{u}]}_{3}^{8} + \frac{1}{4} {[\frac{\sqrt{x^{2} - 4 x + 3}}{x - 2}]}_{- 1}^{0}

= - \frac{1}{2} (4 \sqrt{2} - 2 \sqrt{3}) + \frac{1}{4} (- \frac{\sqrt{3}}{2} - 0)

You can't use 'macro parameter character #' in math mode

Commented by ZiYangLee last updated on 17/Sep/21

Does anyone get the same answer

with me . . ?

Answered by ARUNG_Brandon_MBU last updated on 17/Sep/21

I = \int_{- 1}^{0} \frac{x - 1}{\sqrt{x^{2} - 4 x + 3}} d x

= \frac{1}{2} \int_{- 1}^{0} \frac{(2 x - 4)}{\sqrt{x^{2} - 4 x + 3}} d x + \int_{- 1}^{0} \frac{d x}{\sqrt{x^{2} - 4 x + 3}}

= \frac{1}{2} \int_{- 1}^{0} \frac{d (x^{2} - 4 x + 3)}{\sqrt{x^{2} - 4 x + 3}} + \int_{- 1}^{0} \frac{d x}{\sqrt{{(x - 2)}^{2} - 1}}

= \frac{1}{2} \cdot \frac{2}{1} {[\sqrt{x^{2} - 4 x + 3}]}_{- 1}^{0} + {[argch (x - 2)]}_{- 1}^{0}

= \sqrt{3} - 2 \sqrt{2} + {[\ln ∣ (x - 2) + \sqrt{{(x - 2)}^{2} - 1} ∣]}_{- 1}^{0}

= \sqrt{3} - 2 \sqrt{2} + \ln (\frac{\sqrt{3} - 2}{- 1}) = \sqrt{3} - 2 \sqrt{2} + \ln (2 - \sqrt{3})

Answered by peter frank last updated on 18/Sep/21

Q [152270]