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# Question-77335

Question Number 77335 by aliesam last updated on 05/Jan/20
Commented by aliesam last updated on 05/Jan/20
Commented by msup trace by abdo last updated on 05/Jan/20
$$\int\:\frac{{x}^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx}\:\:={I}\:\Rightarrow \\$$$${I}=\int\:\frac{{x}^{\mathrm{3}} +\mathrm{1}−\mathrm{2}}{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx}\:={x}−\mathrm{2}\:\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\$$$${let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)} \\$$$${F}\left({x}\right)=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\$$$${a}=\left({x}+\mathrm{1}\right){F}\left({x}\right)\mid_{{x}=−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}} \\$$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{3}} \\$$$${F}\left(\mathrm{0}\right)=\mathrm{1}\:={a}+{c}\:\Rightarrow{c}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\$$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}+\frac{\mathrm{2}}{\mathrm{3}}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\$$$$\Rightarrow\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid \\$$$$−\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\$$$${now}\:{its}\:{eazy}\:{to}\:{solve}… \\$$$$\\$$
Commented by aliesam last updated on 05/Jan/20
$${perfect}\:{sir}\:{thank}\:{you} \\$$
Commented by mathmax by abdo last updated on 05/Jan/20
$${you}\:{are}\:{welcome}. \\$$
Answered by petrochengula last updated on 05/Jan/20
$$=\int\frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} +\mathrm{1}}{dx}−\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\$$$$=\int\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}\right){dx}−\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\$$$$=\int{dx}−\mathrm{2}\int\frac{\mathrm{1}}{{x}^{\mathrm{3}} +\mathrm{1}}{dx} \\$$$$={x}−\mathrm{2}\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx} \\$$$$={x}−\mathrm{2}\left(\int\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}{dx}+\int\frac{\mathrm{2}−{x}}{\mathrm{3}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}\right) \\$$$$={x}−\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{x}−\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\$$$${I}\:{do}\:{hope}\:{you}\:{will}\:{be}\:{able}\:{to}\:{continue}\:{from}\:{here} \\$$
Commented by aliesam last updated on 05/Jan/20
$${thank}\:{hou}\:{sir} \\$$