Question Number 13977 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
Commented by prakash jain last updated on 26/May/17
$$\int\frac{\mathrm{3}+{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{3}} }{dx} \\ $$$$\int\frac{\mathrm{3}}{\left(\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)^{\mathrm{3}} +{x}^{\mathrm{3}} }{dx}+\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{3}} }{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{3}} }{dx}=\int\frac{{du}}{\mathrm{3}\left(\mathrm{2}+{u}\right)}\:\:\:\:\left({x}^{\mathrm{3}} ={u}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{2}+{u}\right)+{C}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{2}+{x}^{\mathrm{3}} \right) \\ $$$$−−−−−− \\ $$$$\left({x}^{\mathrm{3}} +\mathrm{2}\right)=\left({x}+\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)\left({x}+\mathrm{2}^{\mathrm{1}/\mathrm{3}} {w}\right)\left({x}+\mathrm{2}^{\mathrm{1}/\mathrm{3}} {w}^{\mathrm{2}} \right) \\ $$$$\left(\mathrm{1},{w},{w}^{\mathrm{2}} \right)\:\mathrm{are}\:\mathrm{cube}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{unity}. \\ $$$${for}\:{easier}\:{typing}\:{i}\:{will}\:{put}\:\mathrm{2}^{\mathrm{1}/\mathrm{3}} ={j} \\ $$$$\int\frac{\mathrm{3}{dx}}{\left({x}+{j}\right)\left({x}+{jw}\right)\left({x}+{jw}^{\mathrm{2}} \right)} \\ $$$$=\int\frac{{A}}{{x}+{j}}{dx}+\int\frac{{B}}{{x}+{jw}}{dx}+\int\frac{{C}}{{x}+{jw}^{\mathrm{2}} }{dx} \\ $$$${perform}\:{partial}\:{fraction}\:{and} \\ $$$${integrate}. \\ $$
Answered by ajfour last updated on 26/May/17
![I=∫_0 ^( 1) (((x^2 +3)/(x^3 +2)))dx let a^3 =2 x^3 +2=(x+a)(x^2 −ax+a^2 ) for x=1, (1+a)(1−a+a^2 )=3 ((x^2 +3)/(x^3 +2))=(A/(x+a))+((Bx+C)/(x^2 −ax+a^2 )) A=(1/3)+(a/2), B=(2/3)−(a/2), C=(2/a)−(a/3) I=Aln (1+(1/a))+(B/2)∫_0 ^( 1) (((2x−a)dx)/(x^2 −ax+a^2 )) +(C+((aB)/2))∫_0 ^( 1) (dx/((x−(a/2))^2 +(((a(√3))/2))^2 )) I =Aln (1+(1/a))+(B/2)ln (((1−a+a^2 )/a^2 )) +(C+((aB)/2))(2/(a(√3)))[tan^(−1) (((a^2 −1)/(√3)))+(π/6)] substiting for A,B, C and utilizing the fact a^2 =(2/a), that a^2 −a+1=(3/(a+1)) we have I=((3a)/4)ln (1+a)−((a/4)+(2/3))ln a+ ((1/3)−(a/4))ln 3+((a(√3))/2)[tan^(−1) (((a^2 −1)/(√3)))+(π/6)] I=((3a)/4)ln (1+(1/a))+((a(√3))/2)tan^(−1) (((√3)/(2a−1))) −(a/4)ln (3a)+(2/3)ln (((√3)/a)) .](Q14012.png)
$$\:{I}=\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{3}} +\mathrm{2}}\right){dx} \\ $$$$\:\:{let}\:{a}^{\mathrm{3}} =\mathrm{2} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}=\left({x}+{a}\right)\left({x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} \right) \\ $$$${for}\:{x}=\mathrm{1},\:\:\left(\mathrm{1}+{a}\right)\left(\mathrm{1}−{a}+{a}^{\mathrm{2}} \right)=\mathrm{3} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{3}}{{x}^{\mathrm{3}} +\mathrm{2}}=\frac{{A}}{{x}+{a}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} } \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{3}}+\frac{{a}}{\mathrm{2}},\:\:{B}=\frac{\mathrm{2}}{\mathrm{3}}−\frac{{a}}{\mathrm{2}},\:{C}=\frac{\mathrm{2}}{{a}}−\frac{{a}}{\mathrm{3}} \\ $$$${I}={A}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)+\frac{{B}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{\left(\mathrm{2}{x}−{a}\right){dx}}{{x}^{\mathrm{2}} −{ax}+{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({C}+\frac{{aB}}{\mathrm{2}}\right)\int_{\mathrm{0}} ^{\:\:\mathrm{1}} \frac{{dx}}{\left({x}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\:{I}\:={A}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)+\frac{{B}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{1}−{a}+{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right) \\ $$$$\:+\left({C}+\frac{{aB}}{\mathrm{2}}\right)\frac{\mathrm{2}}{{a}\sqrt{\mathrm{3}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{6}}\right] \\ $$$$\:\:{substiting}\:{for}\:{A},{B},\:{C}\:{and} \\ $$$${utilizing}\:{the}\:{fact}\:\:{a}^{\mathrm{2}} =\frac{\mathrm{2}}{{a}}, \\ $$$${that}\:\:{a}^{\mathrm{2}} −{a}+\mathrm{1}=\frac{\mathrm{3}}{{a}+\mathrm{1}}\:\:{we}\:{have} \\ $$$${I}=\frac{\mathrm{3}{a}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{1}+{a}\right)−\left(\frac{{a}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{ln}\:{a}+ \\ $$$$\:\:\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{{a}}{\mathrm{4}}\right)\mathrm{ln}\:\mathrm{3}+\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} −\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\pi}{\mathrm{6}}\right] \\ $$$${I}=\frac{\mathrm{3}{a}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)+\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}{a}−\mathrm{1}}\right) \\ $$$$\:\:\:\:\:−\frac{{a}}{\mathrm{4}}\mathrm{ln}\:\left(\mathrm{3}{a}\right)+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\:\left(\frac{\sqrt{\mathrm{3}}}{{a}}\right)\:. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/May/17
$${thank}\:{you}\:{so}\:{much}\:{dear}\:{mr}\:{ajfour}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/May/17
![A=(1/(2)^(1/3) )⇒t=Ax (x/(2)^(1/3) )=t⇒(x/(√3))=((t(2)^(1/3) )/(√3))=mt,m=((2)^(1/3) /(√3)),dx=(2)^(1/3) dt=m(√3)dt I=∫((3(1+((x/(√3)))^2 ))/(2(1+((x/(2)^(1/3) ))^3 )))dx=((3m(√3))/2)∫((1+mt^2 )/(1+t^3 ))dt =((3m(√3))/2)[∫(dt/(1+t^3 ))+m∫((t^2 dt)/(1+t^3 ))]=A[I_1 +mI_2 ] 1)I_2 =∫((t^2 dt)/(1+t^3 ))=(1/3)ln(1+t^3 )+C_2 2)I_1 =∫(dt/(1+t^3 ))=? (1/(1+t^3 ))=(a/(1+t))+((bx+c)/(1−t+t^2 )) { ((t=0⇒1=a+c)),((t=1⇒(1/2)=(a/2)+b+c)) :} t=−1⇒(1/3)=a+0⇒(a=(1/3),c=(2/3),b=((−1)/3)) ⇒(1/(1+t^3 ))=(1/(3(1+t)))+((−t+2)/(3(1−t+t^2 ))) ⇒I_1 =(1/3)∫((1/(1+t))+((−t+2)/(t^2 −t+1)))dt= =(1/3)∫((1/(1+t)))dt−(1/6)∫(((2t−1)−3)/(t^2 −t+1))dt= =(1/3)∫(dt/(1+t))−(1/6)∫(( 2t−1)/(t^2 −t+1))dt+((√3)/4)∫(dt/((((2t−1)/(√3)))^2 +1))= =(1/3)ln(1+t)−(1/6)ln(1−t+t^2 )+((√3)/4)tg^(−1) (((2t−1)/(√3)))+C_1 I=((3(2)^(1/3) )/2)[(1/3)ln(1+t)−(1/6)ln(1−t+t^2 )+↓ +((√3)/4)tg^(−1) (((2t−1)/(√3)))+((2)^(1/3) /(√3)).(1/3)ln(1+t^3 )]+C =(1/(8A))[4ln(1+Ax)−2ln(1−Ax+A^2 x^2 )+ +3(√3)tg^(−1) (((2Ax−1)/(√3)))+((4(√3))/(3A))ln(1+A^3 x^3 )]+C. F(1)=((2)^(1/3) /8)[4ln(1+(1/(2)^(1/3) ))−2ln(1−(1/(2)^(1/3) )+(1/(4)^(1/3) ))+ +3(√3)tg^(−1) ((((4)^(1/3) −1)/(√3)))+((4(2)^(1/3) )/(√3))ln(3/2)] F(0)=3(√3)tg^(−1) (−(1/(√3)))=−((π(√3))/2) . I=F(1)−F(0)=F(1)+((π(√3))/2) .](Q14027.png)
$${A}=\frac{\mathrm{1}}{\sqrt[{\mathrm{3}}]{\mathrm{2}}}\Rightarrow{t}={Ax} \\ $$$$\frac{{x}}{\sqrt[{\mathrm{3}}]{\mathrm{2}}}={t}\Rightarrow\frac{{x}}{\sqrt{\mathrm{3}}}=\frac{{t}\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\sqrt{\mathrm{3}}}={mt},{m}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\sqrt{\mathrm{3}}},{dx}=\sqrt[{\mathrm{3}}]{\mathrm{2}}{dt}={m}\sqrt{\mathrm{3}}{dt} \\ $$$${I}=\int\frac{\mathrm{3}\left(\mathrm{1}+\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right)}{\mathrm{2}\left(\mathrm{1}+\left(\frac{{x}}{\sqrt[{\mathrm{3}}]{\mathrm{2}}}\right)^{\mathrm{3}} \right)}{dx}=\frac{\mathrm{3}{m}\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{\mathrm{1}+{mt}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{3}} }{dt} \\ $$$$=\frac{\mathrm{3}{m}\sqrt{\mathrm{3}}}{\mathrm{2}}\left[\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }+{m}\int\frac{{t}^{\mathrm{2}} {dt}}{\mathrm{1}+{t}^{\mathrm{3}} }\right]={A}\left[{I}_{\mathrm{1}} +{mI}_{\mathrm{2}} \right] \\ $$$$\left.\mathrm{1}\right){I}_{\mathrm{2}} =\int\frac{{t}^{\mathrm{2}} {dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)+{C}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){I}_{\mathrm{1}} =\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=? \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{3}} }=\frac{{a}}{\mathrm{1}+{t}}+\frac{{bx}+{c}}{\mathrm{1}−{t}+{t}^{\mathrm{2}} }\begin{cases}{{t}=\mathrm{0}\Rightarrow\mathrm{1}={a}+{c}}\\{{t}=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}=\frac{{a}}{\mathrm{2}}+{b}+{c}}\end{cases} \\ $$$${t}=−\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}={a}+\mathrm{0}\Rightarrow\left({a}=\frac{\mathrm{1}}{\mathrm{3}},{c}=\frac{\mathrm{2}}{\mathrm{3}},{b}=\frac{−\mathrm{1}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{t}\right)}+\frac{−{t}+\mathrm{2}}{\mathrm{3}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}+\frac{−{t}+\mathrm{2}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{\mathrm{1}+{t}}\right){dt}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\left(\mathrm{2}{t}−\mathrm{1}\right)−\mathrm{3}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{\mathrm{1}+{t}}−\frac{\mathrm{1}}{\mathrm{6}}\int\frac{\:\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{dt}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\int\frac{{dt}}{\left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}\right)−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{tg}^{−\mathrm{1}} \left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+{C}_{\mathrm{1}} \\ $$$${I}=\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}\right)−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)+\downarrow\right. \\ $$$$\left.+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{tg}^{−\mathrm{1}} \left(\frac{\mathrm{2}{t}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\sqrt{\mathrm{3}}}.\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\right]+\boldsymbol{{C}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{A}}\left[\mathrm{4}{ln}\left(\mathrm{1}+{Ax}\right)−\mathrm{2}{ln}\left(\mathrm{1}−{Ax}+{A}^{\mathrm{2}} {x}^{\mathrm{2}} \right)+\right. \\ $$$$\left.+\mathrm{3}\sqrt{\mathrm{3}}{tg}^{−\mathrm{1}} \left(\frac{\mathrm{2}{Ax}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}{A}}{ln}\left(\mathrm{1}+{A}^{\mathrm{3}} {x}^{\mathrm{3}} \right)\right]+\boldsymbol{{C}}. \\ $$$${F}\left(\mathrm{1}\right)=\frac{\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\mathrm{8}}\left[\mathrm{4}{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\sqrt[{\mathrm{3}}]{\mathrm{2}}}\right)−\mathrm{2}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt[{\mathrm{3}}]{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt[{\mathrm{3}}]{\mathrm{4}}}\right)+\right. \\ $$$$\left.+\mathrm{3}\sqrt{\mathrm{3}}{tg}^{−\mathrm{1}} \left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{4}}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{4}\sqrt[{\mathrm{3}}]{\mathrm{2}}}{\sqrt{\mathrm{3}}}{ln}\frac{\mathrm{3}}{\mathrm{2}}\right] \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{3}\sqrt{\mathrm{3}}{tg}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)=−\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}\:. \\ $$$${I}={F}\left(\mathrm{1}\right)−{F}\left(\mathrm{0}\right)={F}\left(\mathrm{1}\right)+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:. \\ $$$$ \\ $$