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Question Number 48719 by Abdo msup. last updated on 27/Nov/18

find ∫ ((x−2)/(√(x^2 +4x−3)))dx

$${find}\:\:\int\:\:\:\:\frac{{x}−\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}}}{dx} \\ $$

Commented by Abdo msup. last updated on 27/Nov/18

I=∫ ((x−2)/(√(x^2 +4x+4−7))) =∫ ((x−2)/(√((x+2)^2 −7)))dx =_(x+2=(√7)ch(t)) ∫ (((√7)ch(t)−4)/((√7)sh(t))) (√7)sh(t)dt =(√7)∫ ch(t) −4t +c =(√7)sh(t)−4t +c but sh(t)=((e^t −e^(−t) )/2) and t=argch(((x+2)/(√7))) =ln(x+(√((((x+2)/(√7)))^2 −1))) ⇒sh(t)=((x+(√((((x+2)/((√7) )))^2 −1))−(1/(x+(√((((x+2)/(√7)))−1)))))/2) I = ((√7)/2){ (x+(√((((x+2)/(√7)))^2 −1))−(x+(√((((x+2)/(√7)))^2 −1)))^(−1) }−4t +c .

$${I}=\int\:\:\frac{{x}−\mathrm{2}}{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}{x}+\mathrm{4}−\mathrm{7}}}\:=\int\:\:\frac{{x}−\mathrm{2}}{\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{7}}}{dx} \\ $$$$=_{{x}+\mathrm{2}=\sqrt{\mathrm{7}}{ch}\left({t}\right)} \:\:\:\:\int\:\:\:\frac{\sqrt{\mathrm{7}}{ch}\left({t}\right)−\mathrm{4}}{\sqrt{\mathrm{7}}{sh}\left({t}\right)}\:\sqrt{\mathrm{7}}{sh}\left({t}\right){dt}\: \\ $$$$=\sqrt{\mathrm{7}}\int\:\:{ch}\left({t}\right)\:−\mathrm{4}{t}\:+{c}\:\:=\sqrt{\mathrm{7}}{sh}\left({t}\right)−\mathrm{4}{t}\:+{c}\:{but} \\ $$$${sh}\left({t}\right)=\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}\:\:{and}\:{t}={argch}\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}}\right) \\ $$$$={ln}\left({x}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow{sh}\left({t}\right)=\frac{{x}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}\:}\right)^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{x}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}}\right)−\mathrm{1}}}}{\mathrm{2}} \\ $$$${I}\:=\:\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\left\{\:\left({x}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}}\right)^{\mathrm{2}} −\mathrm{1}}−\left({x}+\sqrt{\left(\frac{{x}+\mathrm{2}}{\sqrt{\mathrm{7}}}\right)^{\mathrm{2}} −\mathrm{1}}\right)^{−\mathrm{1}} \right\}−\mathrm{4}{t}\:+{c}\:.\right. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

t^2 =x^2 +4x−3 2t(dt/dx)=2x+4 tdt=(x+2)dx ∫((x+2−4)/(√(x^2 +4x−3)))dx ∫((tdt)/t)−4∫(dx/(√((x+2)^2 −((√7) )^2 ))) t−4ln{(x+2)+(√((x+2)^2 −((√7) )^2 )) }+c (√((x^2 +4x−3))) −4ln{(x+2)+(√(x^2 +4x−3)) }+c

$${t}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3} \\ $$$$\mathrm{2}{t}\frac{{dt}}{{dx}}=\mathrm{2}{x}+\mathrm{4} \\ $$$${tdt}=\left({x}+\mathrm{2}\right){dx} \\ $$$$\int\frac{{x}+\mathrm{2}−\mathrm{4}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}}}{dx} \\ $$$$\int\frac{{tdt}}{{t}}−\mathrm{4}\int\frac{{dx}}{\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{7}}\:\right)^{\mathrm{2}} }} \\ $$$${t}−\mathrm{4}{ln}\left\{\left({x}+\mathrm{2}\right)+\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{7}}\:\right)^{\mathrm{2}} }\:\right\}+{c} \\ $$$$\sqrt{\left({x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}\right)}\:−\mathrm{4}{ln}\left\{\left({x}+\mathrm{2}\right)+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}}\:\right\}+{c} \\ $$

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