What is the integral of $\int (x^{2}) \cdot e^{x^{2}} d x$ $int (x^2)*e^(x^2) dx$ ?

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Answer 1 · 2016-09-24T19:30:31

$\frac{1}{2} x e^{x^{2}} - \frac{\sqrt{π}}{4} erfi (x)$ $1/2xe^(x^2)-sqrt(pi)/4"erfi"(x)$

$\frac{d}{d x} (x e^{x^{2}}) = 2 x^{2} e^{x^{2}} + e^{x^{2}}$ $d/dx(xe^(x^2))= 2x^2e^(x^2)+e^(x^2)$

then

$\int x^{2} e^{x^{2}} d x = \frac{1}{2} (x e^{x^{2}} - \int e^{x^{2}} d x)$ $int x^2 e^(x^2)dx = 1/2(xe^(x^2)-int e^(x^2)dx)$

but $erfi (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{x^{2}} d x$ $"erfi"(x) = 2/sqrt(pi)int_0^x e^(x^2)dx$ so

$\int x^{2} e^{x^{2}} d x = \frac{1}{2} x e^{x^{2}} - \frac{\sqrt{π}}{4} erfi (x)$ $int x^2 e^(x^2)dx = 1/2xe^(x^2)-sqrt(pi)/4"erfi"(x)$

NOTE:

$erfi (x)$ $"erfi"(x)$ or error function is defined as

$erfi (x) = \frac{2}{\sqrt{π}} \int_{0}^{x} e^{- ξ^{2}} d ξ$ $"erfi"(x) = 2/sqrtpi int_0^x e^(-xi^2) d xi$

What is the integral of int (x^2)*e^(x^2) dx ∫(x2)⋅ex2dxint (x^2)*e^(x^2) dx ?