Integrate $xe^{-c^2 x^2} \, dx$, or solve the $dv$ in integration by parts

Question

Integrate $xe^{-c^2 x^2} \, dx$. I tried to do it by parts but i dont know how to integrate $dv = e^{-c^2 x^2}\, dx$. Can someone help to either integrate the full or at least the $dv$ because then i can figure it out from there

Answer 1

$$\int x e^{\large -c^2x^2}\,dx = \int e^{\large \overbrace{-c^2x^2}^{\large u}}\,\color{blue}{(x\,dx)}$$

No need for integration by parts. You can do this integral by using substitution.

Put $\,u = -(c^2)\,x^2$. $$u = -(c^2)x^2 \implies du = -2c^2\color{blue}{x\,dx} \iff \dfrac{du}{-2c^2} = \color{blue}{x\,dx}$$

That gives us $$\begin{align} -\dfrac{1}{2c^2}\int e^u\,du & = -\dfrac{1}{2c^2}e^u + C \\ \\ & = -\dfrac{e^{\large -c^2x^2}}{2c^2} + C\end{align}$$

Answer 2

You can use the substitution $ y = -c^2 x^2 $ and you'll be left with a standard integral in terms of $y$. Afterwards, you can revert back to $x$ using your substitution :)

Hint:

$ y = -c^2 x^2 \implies \dfrac{dy}{dx} = -2c^2 x \implies x dx = \dfrac{dy}{-2c^2} $