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Question Number 85866 by liki last updated on 25/Mar/20

Commented by liki last updated on 25/Mar/20

.....Help me plz

$$.....{Help}\:{me}\:{plz} \\ $$

Answered by TANMAY PANACEA. last updated on 25/Mar/20

no easy way... ∫x^4 e^x^2 dx ∫x^4 (1+x^2 +(x^4 /(2!))+(x^6 /(3!))+..)dx e^t =1+t+(t^2 /(2!))+(t^3 /(3!))+(t^4 /(4!))+... ∫(x^4 +(x^6 /1)+(x^8 /(2!))+(x^(10) /(3!))+...) dx =(x^5 /5)+(x^7 /7)+(x^9 /(2!×9))+(x^(11) /(3!×11))+...

$${no}\:{easy}\:{way}... \\ $$$$\int{x}^{\mathrm{4}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\int{x}^{\mathrm{4}} \left(\mathrm{1}+{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+..\right){dx} \\ $$$${e}^{{t}} =\mathrm{1}+{t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{t}^{\mathrm{4}} }{\mathrm{4}!}+... \\ $$$$\int\left({x}^{\mathrm{4}} +\frac{{x}^{\mathrm{6}} }{\mathrm{1}}+\frac{{x}^{\mathrm{8}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{10}} }{\mathrm{3}!}+...\right)\:{dx} \\ $$$$=\frac{{x}^{\mathrm{5}} }{\mathrm{5}}+\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+\frac{{x}^{\mathrm{9}} }{\mathrm{2}!×\mathrm{9}}+\frac{{x}^{\mathrm{11}} }{\mathrm{3}!×\mathrm{11}}+... \\ $$

Commented by liki last updated on 25/Mar/20

...thank you very much sir.

$$...{thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$

Answered by niroj last updated on 25/Mar/20

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