Question Number 88131 by M±th+et£s last updated on 08/Apr/20
$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}−\sqrt{\mathrm{1}−\mathrm{2}{x}}}{dx} \\ $$
Answered by MJS last updated on 08/Apr/20
![∫((x^2 +1)/(x−(√(1−2x))))dx= [t=(√(1−2x)) → dx−(√(1−2x))dt] =(1/2)∫((t(t^4 −2t^2 +5))/(t^2 +2t−1))dt= =(6−4(√2))∫(dt/(t+1−(√2)))+(6+4(√2))∫(dt/(t+1+(√2)))+∫((t^3 /2)−t^2 +((3t)/2)−4)dt= =(6−4(√2))ln (t+1−(√2)) +(6+4(√2))ln (t+1+(√2)) +(t^4 /8)−(t^3 /3)+((3t^2 )/4)−4t= =(6−4(√2))ln (1−(√2)+(√(1−2x))) + +(6+4(√2))ln (1+(√2)+(√(1−2x))) + +(((2x−7)(2x−1))/8)−((2x−13)/3)(√(1−2x)) +C](Q88139.png)
$$\int\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}−\sqrt{\mathrm{1}−\mathrm{2}{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{1}−\mathrm{2}{x}}\:\rightarrow\:{dx}−\sqrt{\mathrm{1}−\mathrm{2}{x}}{dt}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}\left({t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} +\mathrm{5}\right)}{{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}}{dt}= \\ $$$$=\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}\right)\int\frac{{dt}}{{t}+\mathrm{1}−\sqrt{\mathrm{2}}}+\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\int\frac{{dt}}{{t}+\mathrm{1}+\sqrt{\mathrm{2}}}+\int\left(\frac{{t}^{\mathrm{3}} }{\mathrm{2}}−{t}^{\mathrm{2}} +\frac{\mathrm{3}{t}}{\mathrm{2}}−\mathrm{4}\right){dt}= \\ $$$$=\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}\right)\mathrm{ln}\:\left({t}+\mathrm{1}−\sqrt{\mathrm{2}}\right)\:+\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\mathrm{ln}\:\left({t}+\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{{t}^{\mathrm{4}} }{\mathrm{8}}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{4}{t}= \\ $$$$=\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}\right)\mathrm{ln}\:\left(\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{1}−\mathrm{2}{x}}\right)\:+ \\ $$$$\:\:\:\:\:+\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{1}−\mathrm{2}{x}}\right)\:+ \\ $$$$\:\:\:\:\:+\frac{\left(\mathrm{2}{x}−\mathrm{7}\right)\left(\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{8}}−\frac{\mathrm{2}{x}−\mathrm{13}}{\mathrm{3}}\sqrt{\mathrm{1}−\mathrm{2}{x}}\:+{C} \\ $$
Commented by M±th+et£s last updated on 08/Apr/20
$${nice}\:{solution}\:{sir} \\ $$