Question Number 40145 by maxmathsup by imad last updated on 16/Jul/18
$${calculate}\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\frac{\left({x}−\mathrm{1}\right){dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}} \\ $$
Commented by math khazana by abdo last updated on 18/Jul/18
![let I = ∫_(−7) ^(−3) (((x−1)dx)/( (√(x^2 +2x −3)))) we have I = (1/2) ∫_(−7) ^(−3) ((2x+2−4)/( (√(x^2 +2x−3))))dx =[ (√(x^2 +2x−3))]_(−7) ^(−3) −2 ∫_(−7) ^(−3) (dx/( (√(x^2 +2x−3)))) =(√(9−6−3)) −(√(49−17)) −2 ∫_(−7) ^(−3) (dx/( (√(x^2 +2x−3)))) =−(√(32)) −2∫_(−7) ^(−3) (dx/( (√(x^2 +2x−3)))) but ∫_(−7) ^(−3) (dx/( (√(x^2 +2x−3)))) =∫_(−7) ^(−3) (dx/( (√((x+1)^2 −4)))) =_(x+1=2u) ∫_(−3) ^(−1) ((2du)/(2(√(u^2 −1)))) =[ln∣u^2 +(√(u^2 −1))]_(−3) −1 =−ln(9+(√8)) ⇒ I =−4(√2) +2ln(9+2(√2))](https://www.tinkutara.com/question/Q40295.png)
$${let}\:{I}\:=\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\:\frac{\left({x}−\mathrm{1}\right){dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:−\mathrm{3}}}\:\:{we}\:{have} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\frac{\mathrm{2}{x}+\mathrm{2}−\mathrm{4}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}}{dx} \\ $$$$=\left[\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}\right]_{−\mathrm{7}} ^{−\mathrm{3}} \:\:−\mathrm{2}\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}} \\ $$$$=\sqrt{\mathrm{9}−\mathrm{6}−\mathrm{3}}\:−\sqrt{\mathrm{49}−\mathrm{17}}\:−\mathrm{2}\:\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}} \\ $$$$=−\sqrt{\mathrm{32}}\:\:−\mathrm{2}\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}}\:{but} \\ $$$$\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{2}{x}−\mathrm{3}}}\:=\int_{−\mathrm{7}} ^{−\mathrm{3}} \:\:\:\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}} \\ $$$$=_{{x}+\mathrm{1}=\mathrm{2}{u}} \:\:\:\int_{−\mathrm{3}} ^{−\mathrm{1}} \:\:\frac{\mathrm{2}{du}}{\mathrm{2}\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\left[{ln}\mid{u}^{\mathrm{2}} \:+\sqrt{{u}^{\mathrm{2}} \:−\mathrm{1}}\right]_{−\mathrm{3}} −\mathrm{1} \\ $$$$=−{ln}\left(\mathrm{9}+\sqrt{\mathrm{8}}\right)\:\Rightarrow \\ $$$${I}\:=−\mathrm{4}\sqrt{\mathrm{2}}\:\:+\mathrm{2}{ln}\left(\mathrm{9}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18
![I=∫((x−1)/( (√(x^2 +2x−3))))dx =(1/2)∫((2x+2−4)/( (√((x+1)^2 −4))))dx =(1/2)∫((d(x^2 +2x−3))/( (√(x^2 +2x−3))))dx−2∫(dx/( (√((x+1)^2 −4)))) =(1/2)((√(x^2 +2x−3))/(1/2))−2ln{(x+1)+(√((x+1)^2 −4)) } =∣(√(x^2 +2x−3))−2ln{(x+1)+(√((x+1)^2 −4)) }∣_(−7) ^(−3) =0−(√(32)) −2[ln∣−2∣−ln∣−6+(√(32)) ] −(√(32)) −2ln∣(2/( (√(32)) −6))∣](https://www.tinkutara.com/question/Q40294.png)
$${I}=\int\frac{{x}−\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{2}−\mathrm{4}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}}{dx}−\mathrm{2}\int\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}}{\frac{\mathrm{1}}{\mathrm{2}}}−\mathrm{2}{ln}\left\{\left({x}+\mathrm{1}\right)+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}\:\right\} \\ $$$$=\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}}−\mathrm{2}{ln}\left\{\left({x}+\mathrm{1}\right)+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}}\:\right\}\mid_{−\mathrm{7}} ^{−\mathrm{3}} \\ $$$$=\mathrm{0}−\sqrt{\mathrm{32}}\:−\mathrm{2}\left[{ln}\mid−\mathrm{2}\mid−{ln}\mid−\mathrm{6}+\sqrt{\mathrm{32}}\:\right] \\ $$$$−\sqrt{\mathrm{32}}\:−\mathrm{2}{ln}\mid\frac{\mathrm{2}}{\:\sqrt{\mathrm{32}}\:−\mathrm{6}}\mid \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18
$$\:{i}\:{forgot}\:{to}\:{put}\:{ln}… \\ $$
Commented by math khazana by abdo last updated on 19/Jul/18
$${nvermind}\:{the}\:{train}\:{is}\:{going}… \\ $$