Question Number 81921 by M±th+et£s last updated on 16/Feb/20
Answered by MJS last updated on 16/Feb/20
![∫((√((√x)+(√(x−1))))/( (√x)+1))dx= [t=(√x)+(√(x−1)) → dx=((t^4 −1)/(2t^3 ))dt] =∫(((t−1)(t^2 +1))/( (√t^3 )(t+1)))dt= =−4∫((√t)/(t+1))dt+∫(√t)dt+2∫(dt/( (√t)))−∫(dt/( (√t^3 )))= =−8((√t)−arctan (√t))+(2/3)(√t^3 )+4(√t)+(2/( (√t)))= =(2/3)(√t^3 )−4(√t)+(2/( (√t)))+8arctan (√t) = =((4(x+1+(√x)(√(x−1))−3((√x)+(√(x−1))))/(3(√((√x)+(√(x−1))))))+8arctan (√((√x)+(√(x−1)))) +C](https://www.tinkutara.com/question/Q81931.png)
$$\int\frac{\sqrt{\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}}}{\:\sqrt{{x}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\:\rightarrow\:{dx}=\frac{{t}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{t}^{\mathrm{3}} }{dt}\right] \\ $$$$=\int\frac{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{3}} }\left({t}+\mathrm{1}\right)}{dt}= \\ $$$$=−\mathrm{4}\int\frac{\sqrt{{t}}}{{t}+\mathrm{1}}{dt}+\int\sqrt{{t}}{dt}+\mathrm{2}\int\frac{{dt}}{\:\sqrt{{t}}}−\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{3}} }}= \\ $$$$=−\mathrm{8}\left(\sqrt{{t}}−\mathrm{arctan}\:\sqrt{{t}}\right)+\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{t}^{\mathrm{3}} }+\mathrm{4}\sqrt{{t}}+\frac{\mathrm{2}}{\:\sqrt{{t}}}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{t}^{\mathrm{3}} }−\mathrm{4}\sqrt{{t}}+\frac{\mathrm{2}}{\:\sqrt{{t}}}+\mathrm{8arctan}\:\sqrt{{t}}\:= \\ $$$$=\frac{\mathrm{4}\left({x}+\mathrm{1}+\sqrt{{x}}\sqrt{{x}−\mathrm{1}}−\mathrm{3}\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)\right.}{\mathrm{3}\sqrt{\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}}}+\mathrm{8arctan}\:\sqrt{\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}}\:+{C} \\ $$