calculate ∫_(−7) ^(−3) (((x−1)dx)/(√(x^2 +2x−3)))


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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
$let I =∫_(−7) ^(−3) ((x−1)/(√(x^2 +2x−3)))dx we have x^2 +2x−3 =x^2 +2x+1−4 =(x+1)^2 −4 =(x+1−2)(x+1+2) =(x−1)(x+3) ⇒ I =∫_(−7) ^(−3) ((x−1)/(√((x−1)(x+3))))dx =∫_(−7) ^(−3) ((x−1)/(√((1−x)(−x−3))))dx =−∫_(−7) ^(−3) (((√(1−x)))^2 )/((√(1−x))(√(−x−3))))dx =−∫_(−7) ^(−3) ((√(1−x))/(√(−x−3))) dx we use the changement (√(−x−3))=t ⇒ −x−3=t^2 ⇒−x =3+t^2 ⇒x =−3−t^2 ⇒ I =−∫_2 ^0 ((√(1+3+t^2 ))/t)(−2t)dt =−2∫_0 ^2 (√(t^2 +4)) dt =_(t=2sh(u)) −2∫_0 ^(ln(1+(√2))) 2ch(t)(2ch(t)dt =−8 ∫_0 ^(ln(1+(√2))) ((ch(2t)−1)/2)dt =−4 ∫_0 ^(ln(1+(√2))) (ch(2t)−1)dt=4ln(1+(√2))−4[(1/2)sh(2t)]_0 ^(ln(1+(√2))) =4ln(1+(√2))−2[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(1+(√2))) ⇒ I =4ln(1+(√2))−{(1+(√2))^2 −(1/((1+(√2))^2 ))}$
$l e t I = \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{x^{2} + 2 x - 3}} d x w e h a v e$ $x^{2} + 2 x - 3 = x^{2} + 2 x + 1 - 4 = {(x + 1)}^{2} - 4 = (x + 1 - 2) (x + 1 + 2)$ $= (x - 1) (x + 3) \Rightarrow I = \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{(x - 1) (x + 3)}} d x$ $= \int_{- 7}^{- 3} \frac{x - 1}{\sqrt{(1 - x) (- x - 3)}} d x = - \int_{- 7}^{- 3} \frac{{\sqrt{1 - x})}^{2}}{\sqrt{1 - x} \sqrt{- x - 3}} d x$ $= - \int_{- 7}^{- 3} \frac{\sqrt{1 - x}}{\sqrt{- x - 3}} d x w e u s e t h e c h a n g e m e n t \sqrt{- x - 3} = t \Rightarrow$ $- x - 3 = t^{2} \Rightarrow - x = 3 + t^{2} \Rightarrow x = - 3 - t^{2} \Rightarrow$ $I = - \int_{2}^{0} \frac{\sqrt{1 + 3 + t^{2}}}{t} (- 2 t) d t = - 2 \int_{0}^{2} \sqrt{t^{2} + 4} d t$ $=_{t = 2 s h (u)} - 2 \int_{0}^{l n (1 + \sqrt{2})} 2 c h (t) (2 c h (t) d t = - 8 \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (2 t) - 1}{2} d t$ $= - 4 \int_{0}^{l n (1 + \sqrt{2})} (c h (2 t) - 1) d t = 4 l n (1 + \sqrt{2}) - 4 {[\frac{1}{2} s h (2 t)]}_{0}^{l n (1 + \sqrt{2})}$ $= 4 l n (1 + \sqrt{2}) - 2 {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{l n (1 + \sqrt{2})} \Rightarrow$ $I = 4 l n (1 + \sqrt{2}) - {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}}$
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