Menu Close

x-2-1-x-3-2-dx-




Question Number 35015 by NECx last updated on 14/May/18
∫(x^2 /((1+x^3 )^2 ))dx
$$\int\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} }{dx} \\ $$
Answered by ajfour last updated on 14/May/18
I =(1/3)∫((3x^2 dx)/((1+x^3 )^2 ))       let  t=1+x^3   ⇒ I= (1/3)∫ (dt/t^2 )  = −(1/3)((1/(1+x^3 )))+c           =− (1/(3(1+x^3 ))) .
$${I}\:=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{3}{x}^{\mathrm{2}} {dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{let}\:\:{t}=\mathrm{1}+{x}^{\mathrm{3}} \\ $$$$\Rightarrow\:{I}=\:\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{{dt}}{{t}^{\mathrm{2}} }\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }\right)+{c} \\ $$$$\:\:\:\:\:\:\:\:\:=−\:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}\:. \\ $$
Commented by NECx last updated on 15/May/18
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18
t=1+x^3   dt=3x^2 dx  (1/3)∫t^(−2) dt  =(1/3)×(t^(−1) /(−1))  =((−1)/3)×(1/((1+x^3 )))
$${t}=\mathrm{1}+{x}^{\mathrm{3}} \:\:{dt}=\mathrm{3}{x}^{\mathrm{2}} {dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\int{t}^{−\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{{t}^{−\mathrm{1}} }{−\mathrm{1}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)} \\ $$
Commented by NECx last updated on 15/May/18
Thank you so much
$${Thank}\:{you}\:{so}\:{much} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *