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Question-77335

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

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