Limit of difference quotient of t 4 </msup> 2 </mfrac> + t

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.

Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits
${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}{\left(\frac{x^2+x-1}{8x^2-3}\right)}^{\frac{1}{3}}}$

How do you determine whether the function ${\displaystyle f\left(x\right)=x^2{e}^x}$ is concave up or concave down and its intervals?

Find the limit of ${\displaystyle n^2{\log }^n(1-\frac{c\log n}{n})}$

Find derivatives of the function defined as follows.
${\displaystyle y=e^{-8x}}$

Find the limit:
$\underset{n\to \mathrm{\infty }}{lim}(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})$