Question Number 54259 by 951172235v last updated on 01/Feb/19
Commented by Meritguide1234 last updated on 01/Feb/19
Answered by rahul 19 last updated on 01/Feb/19
$$\mathrm{1}. \\ $$$${I}=\:\int\left({x}^{\mathrm{3}{m}} +{x}^{\mathrm{2}{m}} +{x}^{{m}} \right)\left(\mathrm{2}{x}^{\mathrm{2}{m}} +\mathrm{3}{x}^{{m}} +\mathrm{6}\right)^{\frac{\mathrm{1}}{{m}}} {dx} \\ $$$${I}=\int\left({x}^{\mathrm{3}{m}−\mathrm{1}} +{x}^{\mathrm{2}{m}−\mathrm{1}} +{x}^{{m}−\mathrm{1}} \right)\left(\mathrm{2}{x}^{\mathrm{3}{m}} +\mathrm{3}{x}^{\mathrm{2}{m}} +\mathrm{6}{x}^{{m}} \right)^{\frac{\mathrm{1}}{{m}}} {dx} \\ $$$${put}\:\mathrm{2}{x}^{\mathrm{3}{m}} +\mathrm{3}{x}^{\mathrm{2}{m}} +\mathrm{6}{x}^{{m}} ={t} \\ $$$$\Rightarrow\mathrm{6}{m}\left({x}^{\mathrm{3}{m}−\mathrm{1}} +{x}^{\mathrm{2}{m}−\mathrm{1}} +{x}^{{m}−\mathrm{1}} \right){dx}={dt} \\ $$$$\Rightarrow\:{I}=\frac{\mathrm{1}}{\mathrm{6}{m}}\int\left({t}\right)^{\frac{\mathrm{1}}{\mathrm{m}}} \mathrm{dt} \\ $$$$\Rightarrow{I}=\frac{{m}+\mathrm{1}}{\mathrm{6}}\left(\mathrm{2}{x}^{\mathrm{3}{m}} +\mathrm{3}{x}^{\mathrm{2}{m}} +\mathrm{6}{x}^{{m}} \right)^{\frac{{m}+\mathrm{1}}{{m}}} +{C}. \\ $$
Commented by rahul 19 last updated on 01/Feb/19
thank you sir!��
Commented by Meritguide1234 last updated on 01/Feb/19
$$\mathrm{good}\:\mathrm{rahul} \\ $$