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Question-129903




Question Number 129903 by 0731619177 last updated on 20/Jan/21
Commented by Dwaipayan Shikari last updated on 20/Jan/21
∫e^x^2  dx=Σ_(n=0) ^∞ (1/(n!(2n+1)))x^(2n+1) =(x/1)+(x^3 /3)+(x^5 /(10))+(x^7 /(42))+(x^9 /(216))+...+C  =(x/(1−((x^2 /3)/(1+(x^2 /3)−(((3x^2 )/(10))/(1+((3x^2 )/(10))−((((21)/5)x^2 )/(1+((21)/5)x^2 −..))))))))+C  =(x/(1−(x^2 /(3+x^2 −((9x^2 )/(10+3x^2 −((21x^2 )/(5+21x^2 −..))))))))+C
$$\int{e}^{{x}^{\mathrm{2}} } {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}{x}^{\mathrm{2}{n}+\mathrm{1}} =\frac{{x}}{\mathrm{1}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{10}}+\frac{{x}^{\mathrm{7}} }{\mathrm{42}}+\frac{{x}^{\mathrm{9}} }{\mathrm{216}}+…+{C} \\ $$$$=\frac{{x}}{\mathrm{1}−\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{10}}}{\mathrm{1}+\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{10}}−\frac{\frac{\mathrm{21}}{\mathrm{5}}{x}^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{21}}{\mathrm{5}}{x}^{\mathrm{2}} −..}}}}+{C} \\ $$$$=\frac{{x}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}+{x}^{\mathrm{2}} −\frac{\mathrm{9}{x}^{\mathrm{2}} }{\mathrm{10}+\mathrm{3}{x}^{\mathrm{2}} −\frac{\mathrm{21}{x}^{\mathrm{2}} }{\mathrm{5}+\mathrm{21}{x}^{\mathrm{2}} −..}}}}+{C} \\ $$

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