Antiderivative of $x{e}^{-c{x}^{2}}$

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.

I need to define c in ${\int}_{0}^{\mathrm{\infty}}x{e}^{-c{x}^{2}},$, so that it becomes a probability-mass function (so that it equals 1).

Where do I even begin finding the antiderivative of this? I know the answer will be: $\frac{{e}^{-c{x}^{2}}}{-2c}$.

Trying to use partial integration:

$\int f(x)g(x)dx=f(x)G(x)-\int {f}^{\prime}(x)G(X)$

and picking x as my g(x), and ${e}^{-c{x}^{2}}$ as my f(x) I end up with:

${f}^{\prime}(x)=-2cxf(x)$

$G(X)=\frac{{x}^{2}}{2}$

${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-2cxf(x)\frac{{x}^{2}}{2}dx$

Which simplifies to: ${e}^{-c{x}^{2}}\frac{{x}^{2}}{2}-{\int}_{0}^{\mathrm{\infty}}-c{x}^{4}{e}^{-c{x}^{2}}dx$, which I find is just a mess.