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Question Number 41084 by Tawa1 last updated on 01/Aug/18
∫ (x^3 /(x^6 + 1)) dx
$$\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{6}} \:+\:\mathrm{1}}\:\mathrm{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18
∫(x^3 /((x^2 +1)(x^4 −x^2 +1)))dx t=x^2 dt=2xdx (1/2)∫((tdt)/((t+1)(t^2 −t+1))) (t/((t+1)(t^2 −t+1)))=(p/(t+1))+((qt+r)/(t^2 −t+1)) t=pt^2 −pt+p+qt^2 +rt+qt+r t=t^2 (p+q)+t(−p+q+r)+p+r p+q=0 −p+q+r=1 p+r=0 r+r+r=1 r=(1/3) p=−(1/3) q=(1/3) ∫(((−1)/3)/(t+1))dt+∫(((1/3)t+(1/3))/(t^2 −t+1))dt =((−1)/3)∫(dt/(t+1))+(1/3)∫((t+1)/(t^2 −t+1))dt .((−1)/3)∫(dt/(t+1))+(1/6)∫((2t−1+3)/(t^2 −t+1))dt ((−1)/3)∫(dt/(t+1))+(1/6)∫((2t−1)/(t^2 −t+1))dt+(3/6)∫(dt/(t^2 −2t.(1/2)+(1/4)+1−(1/4))) ((−1)/3)∫(dt/(t+1))+(1/6)∫((d(t^2 −t+1))/(t^2 −t+1))+(3/6)∫(dt/((t−(1/2))^2 +((((√3) )/2))^2 )) ((−1)/3)ln(t+1)+(1/6)ln(t^2 −t+1)+(3/6)×(2/( (√3)))tan^(−1) (((t−(1/2))/(((√3) )/2))) now put x^2 inplace of t ((−1)/3)ln(x^2 +1)+(1/6)ln(x^4 −x^2 +1)+(3/6)×(2/( (√3)))tan^(−2) (((x^2 −(1/2))/(((√3) )/2))
$$\int\frac{{x}^{\mathrm{3}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)}{dx} \\ $$$${t}={x}^{\mathrm{2}} \:\:{dt}=\mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{tdt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)} \\ $$$$\frac{{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}=\frac{{p}}{{t}+\mathrm{1}}+\frac{{qt}+{r}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}} \\ $$$${t}={pt}^{\mathrm{2}} −{pt}+{p}+{qt}^{\mathrm{2}} +{rt}+{qt}+{r} \\ $$$${t}={t}^{\mathrm{2}} \left({p}+{q}\right)+{t}\left(−{p}+{q}+{r}\right)+{p}+{r} \\ $$$${p}+{q}=\mathrm{0} \\ $$$$−{p}+{q}+{r}=\mathrm{1} \\ $$$${p}+{r}=\mathrm{0} \\ $$$${r}+{r}+{r}=\mathrm{1}\:\:{r}=\frac{\mathrm{1}}{\mathrm{3}}\:\:{p}=−\frac{\mathrm{1}}{\mathrm{3}}\:\:{q}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\int\frac{\frac{−\mathrm{1}}{\mathrm{3}}}{{t}+\mathrm{1}}{dt}+\int\frac{\frac{\mathrm{1}}{\mathrm{3}}{t}+\frac{\mathrm{1}}{\mathrm{3}}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$.\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}−\mathrm{1}+\mathrm{3}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}+\frac{\mathrm{3}}{\mathrm{6}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{2}{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{{t}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{6}}\int\frac{{d}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}+\frac{\mathrm{3}}{\mathrm{6}}\int\frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}}{ln}\left({t}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{6}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}}\right) \\ $$$${now}\:{put}\:{x}^{\mathrm{2}} \:\:{inplace}\:{of}\:{t} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{6}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}}\right. \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 01/Aug/18
Wow. God bless you sir
$$\mathrm{Wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18
good night...
$${good}\:{night}… \\ $$

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