Question Number 6945 by Tawakalitu. last updated on 03/Aug/16
$$\int\:\frac{{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }{{x}\:+\:\mathrm{1}}\:\:{dx} \\ $$
Commented by Yozzii last updated on 03/Aug/16
![x=u^3 ⇒dx=3u^2 du x^(2/3) =u^2 ⇒I=∫(x^(2/3) /(x+1))dx=∫((u^2 ×3u^2 )/(u^3 +1))du I=3∫(u^4 /(u^3 +1))du ∵ u−(u/(u^3 +1))=((u^4 +u−u)/(u^3 +1))=(u^4 /(u^3 +1)) ⇒I=3∫(u−(u/(u^3 +1)))du I=3[(u^2 /2)−∫(u/(u^3 +1))du]. (u/(u^3 +1))=(u/((u+1)(u^2 −u+1)))≡(a/(u+1))+((bu+c)/(u^2 −u+1)) ⇒u=a(u^2 −u+1)+(u+1)(bu+c) u=(a+b)u^2 +(c+b−a)u+c+a ⇒a=−b, c=−a, c+b−a=1 ∴ −a−a−a=1⇒a=−1/3 ⇒b=1/3,c=1/3 (u/(u^3 +1))=(1/3)(((u+1)/(u^2 −u+1))−(1/(u+1))) I=3[(u^2 /2)−(1/3)∫(((u+1)/(u^2 −u+1))−(1/(u+1)))du] I=((3u^2 )/2)−∫(((2u−1+3)/(2(u^2 −u+1)))−(1/(u+1)))du I=((3u^2 )/2)−(1/2)∫((2u−1)/(u^2 −u+1))du−(3/2)∫(1/(u^2 −u+1))du+ln∣u+1∣ I=((3u^2 )/2)+ln∣u+1∣−(1/2)ln∣u^2 −u+1∣−(3/2)∫(du/((u−(1/2))^2 +(3/4))) I=((3u^2 )/2)+ln∣u+1∣−(1/2)ln∣u^2 −u+1∣−(3/2)×(1/((√3)/2))tan^(−1) ((u−(1/2))/((√3)/2))+C I=((3u^2 )/2)+ln∣u+1∣−(1/2)ln∣u^2 −u+1∣−(√3)tan^(−1) ((2u−1)/( (√3)))+C u=x^(1/3) ⇒∫(x^(2/3) /(x+1))dx=((3x^(2/3) )/2)+ln∣x^(1/3) +1∣−(1/2)ln∣x^(2/3) −x^(1/3) +1∣−(√3)tan^(−1) ((2x^(1/3) −1)/( (√3)))+C](https://www.tinkutara.com/question/Q6960.png)
$${x}={u}^{\mathrm{3}} \Rightarrow{dx}=\mathrm{3}{u}^{\mathrm{2}} {du} \\ $$$${x}^{\mathrm{2}/\mathrm{3}} ={u}^{\mathrm{2}} \\ $$$$\Rightarrow{I}=\int\frac{{x}^{\mathrm{2}/\mathrm{3}} }{{x}+\mathrm{1}}{dx}=\int\frac{{u}^{\mathrm{2}} ×\mathrm{3}{u}^{\mathrm{2}} }{{u}^{\mathrm{3}} +\mathrm{1}}{du} \\ $$$${I}=\mathrm{3}\int\frac{{u}^{\mathrm{4}} }{{u}^{\mathrm{3}} +\mathrm{1}}{du} \\ $$$$\because\:{u}−\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}=\frac{{u}^{\mathrm{4}} +{u}−{u}}{{u}^{\mathrm{3}} +\mathrm{1}}=\frac{{u}^{\mathrm{4}} }{{u}^{\mathrm{3}} +\mathrm{1}} \\ $$$$\Rightarrow{I}=\mathrm{3}\int\left({u}−\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}\right){du} \\ $$$${I}=\mathrm{3}\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\int\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}{du}\right]. \\ $$$$\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}=\frac{{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)}\equiv\frac{{a}}{{u}+\mathrm{1}}+\frac{{bu}+{c}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}} \\ $$$$\Rightarrow{u}={a}\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)+\left({u}+\mathrm{1}\right)\left({bu}+{c}\right) \\ $$$${u}=\left({a}+{b}\right){u}^{\mathrm{2}} +\left({c}+{b}−{a}\right){u}+{c}+{a} \\ $$$$\Rightarrow{a}=−{b},\:{c}=−{a},\:{c}+{b}−{a}=\mathrm{1} \\ $$$$\therefore\:−{a}−{a}−{a}=\mathrm{1}\Rightarrow{a}=−\mathrm{1}/\mathrm{3} \\ $$$$\Rightarrow{b}=\mathrm{1}/\mathrm{3},{c}=\mathrm{1}/\mathrm{3} \\ $$$$\frac{{u}}{{u}^{\mathrm{3}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right) \\ $$$${I}=\mathrm{3}\left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\int\left(\frac{{u}+\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du}\right] \\ $$$${I}=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}−\int\left(\frac{\mathrm{2}{u}−\mathrm{1}+\mathrm{3}}{\mathrm{2}\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$$${I}=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{u}−\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{1}}{{u}^{\mathrm{2}} −{u}+\mathrm{1}}{du}+{ln}\mid{u}+\mathrm{1}\mid \\ $$$${I}=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}+{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}^{\mathrm{2}} −{u}+\mathrm{1}\mid−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$${I}=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}+{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}^{\mathrm{2}} −{u}+\mathrm{1}\mid−\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{tan}^{−\mathrm{1}} \frac{{u}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}+{C} \\ $$$${I}=\frac{\mathrm{3}{u}^{\mathrm{2}} }{\mathrm{2}}+{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}^{\mathrm{2}} −{u}+\mathrm{1}\mid−\sqrt{\mathrm{3}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{u}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$$${u}={x}^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow\int\frac{{x}^{\mathrm{2}/\mathrm{3}} }{{x}+\mathrm{1}}{dx}=\frac{\mathrm{3}{x}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2}}+{ln}\mid{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}/\mathrm{3}} −{x}^{\mathrm{1}/\mathrm{3}} +\mathrm{1}\mid−\sqrt{\mathrm{3}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}^{\mathrm{1}/\mathrm{3}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 03/Aug/16
$${Woow}….\:{I}\:{love}\:{this}…\:{thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$
Answered by Yozzii last updated on 03/Aug/16
$${An}\:{answer}\:{is}\:{in}\:{comments}. \\ $$