Question Number 73483 by abdomathmax last updated on 13/Nov/19
$${find}\:\int\:\:\:\:\frac{{dx}}{{x}+\mathrm{2}−\sqrt{{x}^{\mathrm{2}} −{x}\:+\mathrm{7}}} \\ $$
Answered by MJS last updated on 13/Nov/19
![∫(dx/(x+2−(√(x^2 −x+7))))=∫(dx/(x+2−(√((x−(1/2))^2 +((27)/4)))))= [t=((√3)/9)(2x−1) → dx=((3(√3))/2)dt] =∫(dt/(t+((5(√3))/9)−(√(t^2 +1))))= [t=sinh ln u =((u^2 −1)/(2u)) ⇒ u=t+(√(t^2 +1)) → dt=((u^2 +1)/(2u^2 ))du] =((3(√3))/2)∫((u^2 +1)/(u(5u−3(√3))))du= =((3(√3))/(10))∫du−(1/2)∫(du/u)+((26)/(25))∫(du/(u−((3(√3))/5)))= =((3(√3))/(10))u−(1/2)ln u +((26)/(25))ln (5u−3(√3)) = ... =((2x−1+2(√(x^2 −x+7)))/(10))−(1/2)ln (2x−1+2(√(x^2 −x+7))) +((26)/(25))ln (5x−16+5(√(x^2 −x+7))) +C](https://www.tinkutara.com/question/Q73517.png)
$$\int\frac{{dx}}{{x}+\mathrm{2}−\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{7}}}=\int\frac{{dx}}{{x}+\mathrm{2}−\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{27}}{\mathrm{4}}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{3}}}{\mathrm{9}}\left(\mathrm{2}{x}−\mathrm{1}\right)\:\rightarrow\:{dx}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}\right] \\ $$$$=\int\frac{{dt}}{{t}+\frac{\mathrm{5}\sqrt{\mathrm{3}}}{\mathrm{9}}−\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sinh}\:\mathrm{ln}\:{u}\:=\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}}\:\Rightarrow\:{u}={t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dt}=\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }{du}\right] \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{{u}\left(\mathrm{5}{u}−\mathrm{3}\sqrt{\mathrm{3}}\right)}{du}= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{10}}\int{du}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{u}}+\frac{\mathrm{26}}{\mathrm{25}}\int\frac{{du}}{{u}−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{5}}}= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{10}}{u}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{u}\:+\frac{\mathrm{26}}{\mathrm{25}}\mathrm{ln}\:\left(\mathrm{5}{u}−\mathrm{3}\sqrt{\mathrm{3}}\right)\:= \\ $$$$… \\ $$$$=\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{7}}}{\mathrm{10}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{7}}\right)\:+\frac{\mathrm{26}}{\mathrm{25}}\mathrm{ln}\:\left(\mathrm{5}{x}−\mathrm{16}+\mathrm{5}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{7}}\right)\:+{C} \\ $$
Commented by abdomathmax last updated on 17/Nov/19
$${thanks}\:{sir}\:{mjs}. \\ $$