Question Number 191243 by sonukgindia last updated on 21/Apr/23
Commented by mr W last updated on 21/Apr/23
$${at}\:{x}=\mathrm{1}\:? \\ $$
Answered by PandaMa last updated on 21/Apr/23
$${at}\:{x}=\mathrm{1}\:{there}\:{is}\:{point}\:{of}\:{singularity}.\:{therefore}\:{that}\:{integral}\:{does}\:{not}\:{converge}.\:{so}\:{we}\:{cannot}\:{calculate}\:{it} \\ $$$$ \\ $$
Answered by mr W last updated on 22/Apr/23
![∫(dx/(x^3 −1)) =∫(dx/((x−1)(x^2 +x+1))) =(1/3)∫[(1/(x−1))−((x+2)/(x^2 +x+1))]dx =(1/3)[ln (x−1)−(1/2)∫(((2x+1)/(x^2 +x+1))+(3/(x^2 +x+1)))dx] =(1/3)[ln (x−1)−(1/2)ln (x^2 +x+1)−(3/2)∫(1/(x^2 +x+1))dx] =(1/3)[ln (x−1)−(1/2)ln (x^2 +x+1)−(3/2)∫(1/((x+(1/4))^2 +(((√3)/2))^2 ))dx] =(1/3)[ln (x−1)−(1/2)ln (x^2 +x+1)−(3/2)×(2/( (√3))) tan^(−1) ((2x+1)/( (√3)))]+C =(1/3)ln (x−1)−(1/6)ln (x^2 +x+1)−(1/( (√3))) tan^(−1) ((2x+1)/( (√3)))+C](https://www.tinkutara.com/question/Q191287.png)
$$\int\frac{{dx}}{{x}^{\mathrm{3}} −\mathrm{1}} \\ $$$$=\int\frac{{dx}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left[\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right]{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\frac{\mathrm{3}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}\right){dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right]+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$