Answer
$x^2 e^x - 2x e^{x}+2e^x- \dfrac{1}{2}e^{x^2}+C$
Work Step by Step
We will solve the given integral by using integrate-by-parts formula such as: $\int udv=uv-\int v du$
$\displaystyle \int (x^2 e^{x}-xe^{x^2}) \ dx=\int x^2 e^{x}\ dx -\int xe^{x^2} \ dx$
or, $=x^2 e^x -\int e^{x} (2x) \ dx- \int xe^{x^2} \ dx$
or, $=x^2 e^x - 2(x e^{x}-e^x)- \dfrac{1}{2}e^{x^2}+C$
or, $=x^2 e^x - 2x e^{x}+2e^x- \dfrac{1}{2}e^{x^2}+C$
