Question Number 66337 by mathmax by abdo last updated on 12/Aug/19
$${calculate}\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\frac{\left({x}−\mathrm{1}\right){dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}} \\ $$
Commented by Prithwish sen last updated on 13/Aug/19
![lim_(ε→−3) ∫_(−7) ^ε ((√(x−1))/( (√(x+3)))) dx = lim_(ε→−3) [(√((x−1)(x+3) ))−4ln∣(√(x+3))+(√(x−1))∣]_(−7) ^ε =4ln∣1+(√2)∣−4(√2) please check.](https://www.tinkutara.com/question/Q66361.png)
$$\mathrm{lim}_{\epsilon\rightarrow−\mathrm{3}} \int_{−\mathrm{7}} ^{\epsilon} \frac{\sqrt{\mathrm{x}−\mathrm{1}}}{\:\sqrt{\mathrm{x}+\mathrm{3}}}\:\mathrm{dx} \\ $$$$=\:\mathrm{lim}_{\epsilon\rightarrow−\mathrm{3}} \:\left[\sqrt{\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\:}−\mathrm{4ln}\mid\sqrt{\mathrm{x}+\mathrm{3}}+\sqrt{\mathrm{x}−\mathrm{1}}\mid\right]_{−\mathrm{7}} ^{\epsilon} \\ $$$$=\mathrm{4ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid−\mathrm{4}\sqrt{\mathrm{2}}\:\:\mathrm{please}\:\mathrm{check}. \\ $$
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 ))}](https://www.tinkutara.com/question/Q66378.png)
$${let}\:{I}\:=\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}}{dx}\:\:{we}\:{have} \\ $$$${x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}\:={x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}−\mathrm{4}\:=\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\:=\left({x}+\mathrm{1}−\mathrm{2}\right)\left({x}+\mathrm{1}+\mathrm{2}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{3}\right)\:\Rightarrow\:{I}\:=\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\frac{{x}−\mathrm{1}}{\:\sqrt{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{3}\right)}}{dx} \\ $$$$=\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\frac{{x}−\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{x}\right)\left(−{x}−\mathrm{3}\right)}}{dx}\:\:=−\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\frac{\left.\sqrt{\mathrm{1}−{x}}\right)^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{x}}\sqrt{−{x}−\mathrm{3}}}{dx} \\ $$$$=−\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\frac{\sqrt{\mathrm{1}−{x}}}{\:\sqrt{−{x}−\mathrm{3}}}\:{dx}\:\:{we}\:{use}\:{the}\:{changement}\:\sqrt{−{x}−\mathrm{3}}={t}\:\Rightarrow \\ $$$$−{x}−\mathrm{3}={t}^{\mathrm{2}} \:\Rightarrow−{x}\:=\mathrm{3}+{t}^{\mathrm{2}} \:\Rightarrow{x}\:=−\mathrm{3}−{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=−\int_{\mathrm{2}} ^{\mathrm{0}} \:\frac{\sqrt{\mathrm{1}+\mathrm{3}+{t}^{\mathrm{2}} }}{{t}}\left(−\mathrm{2}{t}\right){dt}\:=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} \:+\mathrm{4}}\:{dt} \\ $$$$=_{{t}=\mathrm{2}{sh}\left({u}\right)} \:\:\:−\mathrm{2}\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \mathrm{2}{ch}\left({t}\right)\left(\mathrm{2}{ch}\left({t}\right){dt}\:=−\mathrm{8}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\right. \\ $$$$=−\mathrm{4}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \left({ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right){dt}=\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\mathrm{2}\left[\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\Rightarrow \\ $$$${I}\:=\mathrm{4}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\left\{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\} \\ $$$$ \\ $$