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Question-190812




Question Number 190812 by Mahliyo last updated on 12/Apr/23
Commented by Frix last updated on 12/Apr/23
Use 3 steps  1. t=x+(1/2)  2. u=sin^(−1)  ((2t)/( (√5)))  3. v=tan (u/2)
$$\mathrm{Use}\:\mathrm{3}\:\mathrm{steps} \\ $$$$\mathrm{1}.\:{t}={x}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}.\:{u}=\mathrm{sin}^{−\mathrm{1}} \:\frac{\mathrm{2}{t}}{\:\sqrt{\mathrm{5}}} \\ $$$$\mathrm{3}.\:{v}=\mathrm{tan}\:\frac{{u}}{\mathrm{2}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 12/Apr/23
I=∫((xdx)/((1+x)(√(1−x−x^2 ))))     =∫(dx/( (√(1−x−x^2 ))))−∫(dx/((x+1)(√(1−x−x^2 )))), t=(1/(x+1))     =∫(dx/( (√((5/4)−(x+(1/2))^2 ))))+∫(dt/(t(√(1+(1/t)−(1/t^2 )))))     =arcsin(((2x+1)/( (√5))))+∫(dt/( (√(t^2 +t−1))))     =arcsin(((2x+1)/( (√5))))+argch(((2t+1)/( (√5))))+C     =arcsin(((2x+1)/( (√5))))+ln∣((x+3)/( (√5)(x+1)))+((2(√(1−x−x^2 )))/( (√5)(x+1)))∣+C
$${I}=\int\frac{{xdx}}{\left(\mathrm{1}+{x}\right)\sqrt{\mathrm{1}−{x}−{x}^{\mathrm{2}} }} \\ $$$$\:\:\:=\int\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}−{x}^{\mathrm{2}} }}−\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}−{x}−{x}^{\mathrm{2}} }},\:{t}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\:\:\:=\int\frac{{dx}}{\:\sqrt{\frac{\mathrm{5}}{\mathrm{4}}−\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}+\int\frac{{dt}}{{t}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{t}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$\:\:\:=\mathrm{arcsin}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)+\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}} \\ $$$$\:\:\:=\mathrm{arcsin}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{argch}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)+{C} \\ $$$$\:\:\:=\mathrm{arcsin}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{ln}\mid\frac{{x}+\mathrm{3}}{\:\sqrt{\mathrm{5}}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}\left({x}+\mathrm{1}\right)}\mid+{C} \\ $$

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