Question Number 168267 by Florian last updated on 07/Apr/22
$$\:\:\:\:\:\:{Solve}\:\:{this}\:{integral}\:: \\ $$$$\:\:\:\:\:\:\:\int\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{1}} \\ $$$$ \\ $$
Answered by MJS_new last updated on 07/Apr/22
![∫(((√(x+1))−1)/( (√(x+1))+1))dx= [t=(√(x+1)) → dx=2tdt] =2∫ ((t(t−1))/(t+1))dt=∫(2t−4+(4/(t+1)))dt= =t^2 −4t+4ln (t+1) = =x−4(√(x+1))+4ln (1+(√(x+1))) +C](https://www.tinkutara.com/question/Q168273.png)
$$\int\frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt{{x}+\mathrm{1}}+\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}+\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{2}{tdt}\right] \\ $$$$=\mathrm{2}\int\:\frac{{t}\left({t}−\mathrm{1}\right)}{{t}+\mathrm{1}}{dt}=\int\left(\mathrm{2}{t}−\mathrm{4}+\frac{\mathrm{4}}{{t}+\mathrm{1}}\right){dt}= \\ $$$$={t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{4ln}\:\left({t}+\mathrm{1}\right)\:= \\ $$$$={x}−\mathrm{4}\sqrt{{x}+\mathrm{1}}+\mathrm{4ln}\:\left(\mathrm{1}+\sqrt{{x}+\mathrm{1}}\right)\:+{C} \\ $$
Commented by Florian last updated on 07/Apr/22
$${Good}! \\ $$