Question Number 26395 by abdo imad last updated on 25/Dec/17
$${find}\:\:\int\:\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{1}}}\:\: \\ $$
Answered by $@ty@m last updated on 25/Dec/17
$${Let}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$${dx}=\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }{dt}}{\frac{\mathrm{1}}{{t}}\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{t}}−\mathrm{1}}} \\ $$$$\int\frac{\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }{dt}}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\sqrt{\mathrm{1}+{t}−{t}^{\mathrm{2}} }} \\ $$$$=\int\frac{−{dt}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+{t}−{t}^{\mathrm{2}} }} \\ $$$$=\int\frac{−{dt}}{\:\sqrt{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−{t}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{ln}\:\mid\left(\frac{\mathrm{1}}{\mathrm{2}}−{t}\right)+\sqrt{\mathrm{1}+{t}−{t}^{\mathrm{2}} }\mid+{C} \\ $$$$=\mathrm{ln}\:\mid\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{x}}\right)+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\mid+{C} \\ $$