Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# 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##