What is the integral of the error function?

##int”erf”(x)dx = x“erf”(x)+e^(-x^2)/sqrt(pi)+C##
We will use the definition of the error function:
##”erf”(x) = 2/sqrt(pi)int_0^xe^(-t^2)dt##
Along with , , and the .
:
Let ##u = “erf”(x)## and ##dv = dt##
Then, by ,
##du = 2/sqrt(pi)e^(-x^2)## and ##v = x##
By the integration by parts formula ##intudv = uv – intvdu##
##int”erf”(x)dx = x”erf”(x) – int2/sqrt(pi)xe^(-x^2)dx##
:
To evaluate the remaining integral, let ##u = -x^2##
Then ##du = -2xdx## and so
##-int2/sqrt(pi)xe^(-x^2)dx = 1/sqrt(pi)inte^udu##
##=1/sqrt(pi)e^u + C##
##=e^(-x^2)/sqrt(pi)+C##
Putting it all together, we get our final result:
##int”erf”(x)dx = x”erf”(x)+e^(-x^2)/sqrt(pi)+C##