$ \int {(1-x^2)e^{-\frac{x^2}{2}} dx} $

Question

$ \int {(1-x^2)e^{-\frac{x^2}{2}} dx} $

My try,

substituing, $e^{-\frac{x^2}{2}}=y$ but this doesnot work. But, came to know it has very good answer: $xe^{-\frac{x^2}{2}}+c$.

So, if I would substitute $xe^{-\frac{x^2}{2}}=z$ it should work. But this is not legal way.

How to approach this integal in easier way? Any help, suggestion is highly solicited.

Answer 1

What you need to notice to proceed with this integral is partially the solution in fact:

$ \int (1-x^2) e^{\frac{-x^2}{2}}dx = \int e^{\frac{-x^2}{2}} - x^2e^{\frac{-x^2}{2}} dx = \int (x)'e^{\frac{-x^2}{2}} + x (e^{\frac{-x^2}{2}})' dx = \int (xe^{\frac{-x^2}{2}})'dx = xe^{\frac{-x^2}{2}} + c $

Answer 2

Performing integration by parts on the second integral, we have $$ \begin{aligned} I & =\int e^{-\frac{x^2}{2}} d x-\int x^2e^{-\frac{x^2}{2}} d x \\ & =\int e^{-\frac{x^2}{2}} d x+\int x d\left(e^{-\frac{x^2}{2}}\right) \\ & =\int e^{-\frac{x^2}{2}} d x+x e^{-\frac{x^2}{2}}-\int e^{-\frac{x^2}{2}} d x \\ & =x e^{-\frac{x^2}{2}}+C \end{aligned} $$