calculate-7-3-x-1-dx-x-2-2x-3-

Question Number 66337 by mathmax by abdo last updated on 12/Aug/19

c a l c u l a t e \int_{- 7}^{- 3} \frac{(x - 1) d x}{\sqrt{x^{2} + 2 x - 3}}

Commented by Prithwish sen last updated on 13/Aug/19

\lim_{ϵ \to - 3} \int_{- 7}^{ϵ} \frac{\sqrt{x - 1}}{\sqrt{x + 3}} dx

= \lim_{ϵ \to - 3} {[\sqrt{(x - 1) (x + 3)} - 4 \ln ∣ \sqrt{x + 3} + \sqrt{x - 1} ∣]}_{- 7}^{ϵ}

= 4 \ln ∣ 1 + \sqrt{2} ∣ - 4 \sqrt{2} please check .

Commented by mathmax by abdo last updated on 13/Aug/19

l e t I = \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{x^{2} + 2 x - 3}} d x w e h a v e

x^{2} + 2 x - 3 = x^{2} + 2 x + 1 - 4 = {(x + 1)}^{2} - 4 = (x + 1 - 2) (x + 1 + 2)

= (x - 1) (x + 3) \Rightarrow I = \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{(x - 1) (x + 3)}} d x

= \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{(1 - x) (- x - 3)}} d x = - \int_{- 7}^{- 3} \frac{{\sqrt{1 - x})}^{2}}{\sqrt{1 - x} \sqrt{- x - 3}} d x

= - \int_{- 7}^{- 3} \frac{\sqrt{1 - x}}{\sqrt{- x - 3}} d x w e u s e t h e c h a n g e m e n t \sqrt{- x - 3} = t \Rightarrow

- x - 3 = t^{2} \Rightarrow - x = 3 + t^{2} \Rightarrow x = - 3 - t^{2} \Rightarrow

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

=_{t = 2 s h (u)} - 2 \int_{0}^{l n (1 + \sqrt{2})} 2 c h (t) (2 c h (t) d t = - 8 \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (2 t) - 1}{2} d t

= - 4 \int_{0}^{l n (1 + \sqrt{2})} (c h (2 t) - 1) d t = 4 l n (1 + \sqrt{2}) - 4 {[\frac{1}{2} s h (2 t)]}_{0}^{l n (1 + \sqrt{2})}

= 4 l n (1 + \sqrt{2}) - 2 {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{l n (1 + \sqrt{2})} \Rightarrow

I = 4 l n (1 + \sqrt{2}) - {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}}