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
$${thank}\:{hou}\:{sir} \\ $$