Question Number 85866 by liki last updated on 25/Mar/20
Commented by liki last updated on 25/Mar/20
$$…..{Help}\:{me}\:{plz} \\ $$
Answered by TANMAY PANACEA. last updated on 25/Mar/20
$${no}\:{easy}\:{way}… \\ $$$$\int{x}^{\mathrm{4}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\int{x}^{\mathrm{4}} \left(\mathrm{1}+{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+..\right){dx} \\ $$$${e}^{{t}} =\mathrm{1}+{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}+… \\ $$$$\int\left({x}^{\mathrm{4}} +\frac{{x}^{\mathrm{6}} }{\mathrm{1}}+\frac{{x}^{\mathrm{8}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{10}} }{\mathrm{3}!}+…\right)\:{dx} \\ $$$$=\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+\frac{{x}^{\mathrm{9}} }{\mathrm{2}!×\mathrm{9}}+\frac{{x}^{\mathrm{11}} }{\mathrm{3}!×\mathrm{11}}+… \\ $$
Commented by liki last updated on 25/Mar/20
$$…{thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$
Answered by niroj last updated on 25/Mar/20
