My task is to find a recurrence relation for the following definite intergral: $$a_m =\frac{2m+1}{2^{m+1}m!}\int_{-1}^1 (1-x^2)^me^x dx $$ using only $a_{m-1}$ and $a_{m-2}$. I found the first two terms: $$a_0 = \frac{1}{2}(e^1-e^{-1})$$ $$a_1 = 3e^{-1}$$ My attempt at a recurrence relation goes like this. For $a_{m-1}$ and $a_{m-2}$, we have that by definition $$ a_{m-2} =\frac{2m-3}{2^{m-1}(m-2)!}\int_{-1}^1 (1-x^2)e^x dx$$ $$ a_{m-1} =\frac{2m-1}{2^{m+1}(m-1)!}\int_{-1}^1 (1-x^2)^{m-1}e^x dx$$ If we use integration by part on $a_m$, we get $$a_m =\frac{2m+1}{2^{m+1}m!} \{e^x(1-x^2)^m \Big|_{-1}^1 + 2m \int_{-1}^{1} e^x x(1-x^2)^{m-1} \} $$ I know that the first term is clearly 0, so we are left with
$$a_m =\frac{2m+1}{2^{m}(m-1)!} \int_{-1}^{1} e^x x(1-x^2)^{m-1} $$
If we do by parts again, the same way, the first term is again 0 and is
$$a_m =\frac{2m+1}{2^{m}(m-1)!} \{ -\int_{-1}^{1} e^x \Big( (1-x^2)^{m-1} - 2 (m-1)x^2(1-x^2)^{m-2} \Big) \} $$ We can finally use one of the previous terms and $a_m$ becomes
$$a_m = -\frac{2m+1}{2m-1} a_{m-1}+ \frac{2m+1}{2^{m-1}(m-2)!} \int_{-1}^{1} e^x x^2 (1-x^2)^{m-2} \}$$ Which is where I get stuck. I can't seem to get rid of that pesky $x^2$. Any hints would be appreciated, or perhaps a different approach to the problem altogether. Thank you!