Question Number 220850 by fantastic last updated on 20/May/25
$$\int\sqrt{\frac{{x}+\mathrm{1}}{{x}+\mathrm{2}}}.\frac{\mathrm{1}}{{x}+\mathrm{3}}{dx} \\ $$
Commented by Frix last updated on 20/May/25
$$\mathrm{Question}\:\mathrm{219193} \\ $$$$\mathrm{Why}\:\mathrm{you}\:\mathrm{ask}\:\mathrm{again}? \\ $$
Answered by Ghisom last updated on 20/May/25
![∫((√(x+1))/( (x+3)(√(x+2))))dx= [t=((√(x+1))+(√(x+2)))^2 → dx=(((√(x+1))(√(x+2)))/t)dt] =∫(((t−1)^2 )/(t(t^2 +6t+1)))dt=∫(dt/t)−8∫(dt/(t^2 +6t+1))= =ln t +(√2)ln ((t+3+2(√2))/(t+3−2(√2))) = =2ln ((√(x+1))+(√(x+2))) +(√2)ln ((x+3+(√2)+(√(x+1))(√(x+2)))/(x+3−(√2)+(√(x+1))(√(x+2)))) +C](https://www.tinkutara.com/question/Q220940.png)
$$\int\frac{\sqrt{{x}+\mathrm{1}}}{\:\left({x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{2}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right)^{\mathrm{2}} \:\rightarrow\:{dx}=\frac{\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}{{t}}{dt}\right] \\ $$$$=\int\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}\left({t}^{\mathrm{2}} +\mathrm{6}{t}+\mathrm{1}\right)}{dt}=\int\frac{{dt}}{{t}}−\mathrm{8}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{6}{t}+\mathrm{1}}= \\ $$$$=\mathrm{ln}\:{t}\:+\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{{t}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{{t}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}\:= \\ $$$$=\mathrm{2ln}\:\left(\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right)\:+\sqrt{\mathrm{2}}\mathrm{ln}\:\frac{{x}+\mathrm{3}+\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}{{x}+\mathrm{3}−\sqrt{\mathrm{2}}+\sqrt{{x}+\mathrm{1}}\sqrt{{x}+\mathrm{2}}}\:+{C} \\ $$