Integrate e^x^2?

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Answer 1 · 2017-09-05T03:34:38

(i)#int e^(x^2) dx =(sqrt(pi) "erfi(x)")/2 +C#

(ii)#int (e^x)^2 dx = 1/2 e^(2x) +C#

(i) Assuming you mean #e^(x^2)# we will need to use the imaginary error function #"erfi(x)"# defined as:

#"erfi(x)" = 2/sqrt(pi)int_0^x e^(t^2) dt#

Let #I=int e^(x^2) dx#

#I= sqrt(pi)/2 int 2 /sqrt(pi) *e^(x^2)# dx

#=sqrt(pi)/2 int (2 e^(x^2))/sqrt(pi)# dx

#= sqrt(pi)/2 "erfi(x)" +C#

#= (sqrt(pi) "erfi(x)")/2 +C#

(ii) On the other hand, if you meant #(e^x)^2#

Let #I= int (e^x)^2 dx= int e^(2x) dx#

#=1/2e^(2x) +C# [Standard integral]