Question Number 156729 by MathSh last updated on 14/Oct/21
Answered by MJS_new last updated on 15/Oct/21
![∫((x+2)/((x^2 +3x+3)(√(x+1))))dx= [t=(√(x+1)) → dx=2(√(x+1))dt] =2∫((t^2 +1)/(t^4 +t^2 +1))dt=∫(dt/(t^2 −t+1))+∫(dt/(t^2 +t+1))= =((2(√3))/3)arctan (((√3)(2t−1))/3) +((2(√3))/3)arctan (((√3)(2t+1))/3) = =((2(√3))/3)(arctan (((√3)(2(√(x+1))−1))/3) +arctan (((√3)(2(√(x+1))+1))/3))+C](https://www.tinkutara.com/question/Q156738.png)
$$\int\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{3}\right)\sqrt{{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}+\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}+\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\int\frac{{dt}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}+\int\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{3}}\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}}\:= \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}\sqrt{{x}+\mathrm{1}}−\mathrm{1}\right)}{\mathrm{3}}\:+\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}\sqrt{{x}+\mathrm{1}}+\mathrm{1}\right)}{\mathrm{3}}\right)+{C} \\ $$
Commented by MathSh last updated on 15/Oct/21
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thankyou} \\ $$