Calculating: $$I=\int_{-1}^1{(1+x)^{m+n}P_n(x)}dx$$ Where $P_n$ is a Legendre Polynomial.
My progress: Using Rodrigues formula: $$I=\dfrac{1}{2^n n!} \int_{-1}^1{(1+x)^{m+n}\large(\dfrac{d}{dx}\large)^n(x^2-1)^n}dx$$ A first interation by parts leads to: $$I=\dfrac{1}{2^n n!}[(1+x)^{m+n} \large(\dfrac{d}{dx}\large)^{n-1}(x^2-1)^n -\int_{-1}^1{(m+n)(1+x)^{m+n-1}(\dfrac{d}{dx})^{n-1}(x^2-1)^n}dx]$$
The integrated terms vanishes due to $x^2-1$ A second integration with the vanishing terms is: $$I=\dfrac{1}{2^n n!}[\int_{-1}^1{(m+n)(m+n-1)(1+x)^{m+n-2}(\dfrac{d}{dx})^{n-2}(x^2-1)^n}dx]$$
And here is where i have problems, continuing the process $n$-times to vanish the differntial inside the integrand.
$$I=\dfrac{1}{2^n n!}\int_{-1}^1{\dfrac{(m+n)!}{m!}(x^2-1)^n}dx$$
Can´t assure the coefficient is $\dfrac{(m+n)!}{m!}$ How can i finish the evaluation?
I can follow your derivation up to the point where $$ I= \frac{1}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n} \frac{d^n}{dx^n} (x^2-1)^n.$$ I assume for the following that $m,n\in \mathbb{N}_0$ (as was probably intended).
Starting from there, your idea is to integrate by parts ($n$-times). Let us observe what happens after the first application ($n>0$ assumed) $$I = \frac{1}{2^n n!} (1+x)^{m+n} \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \Bigl|_{x=-1}^1 -\frac{m+n}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-1} \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n .$$
Following your argument, we want to discard the boundary term. Now it is easy to check that $d^{n-1}(x^2-1)^n/dx^{n-1}=(x^2-1) p(x)$ with $p(x)$ a polynomial. So the boundary term indeed vanishes.
So given $n>0$, we have that $$I = -\frac{m+n}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-1} \frac{d^{n-1}}{dx^{n-1}} (x^2-1)^n .$$
We continue with the next application of integration by parts (assuming $n>1$) $$I = - \frac{m+n}{2^n n!} (1+x)^{m+n-1} \frac{d^{n-2}}{dx^{n-2}} (x^2-1)^n \Bigl|_{x=-1}^1+ \frac{(m+n)(m+n-1)}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m+n-2} \frac{d^{n-2}}{dx^{n-2}} (x^2-1)^n .$$ This time it is important that $d^{n-2}(x^2-1)^n/dx^{n-2}=(x^2-1)^2 \tilde p(x)$ such that the boundary term vanishes once more.
So we are reasonably sure that after the $n$-th application of integration by parts we have that $$ I = (-1)^n \frac{(m+n)(m+n-1)\cdots (m+1)}{2^n n!} \int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n\\ = \frac{(-1)^n}{2^n} \binom{m+n}{n}\int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n. $$ Of course this result can be proven by induction.
So the remaining task is to evaluate $$J= \int_{-1}^1\!dx\,(1+x)^{m} (x^2-1)^n.$$ Also this is not completely trivial. However, observe that after substitution $2 y= 1+x$, we obtain $$J = 2\int_0^1\!dy\, (2y)^m (-4 (1-y) y)^n = (-1)^n 2^{m+2n+1} \int_0^1\!dy \,y^{m+n} (1-y)^n.$$ The last integral is the definition of the Beta function. Thus we obtain $$J = (-1)^n 2^{m+2n+1} B(m+n+1, n+1) = (-1)^n 2^{m+2n+1} \frac{(m+n)! n!}{(m+2n +1)!}.$$
In conclusion, we have the result $$I =\frac{(-1)^n}{2^n} \binom{m+n}{n}(-1)^n 2^{m+2n+1} \frac{(m+n)!\, n!}{(m+2n +1)!} = \frac{2^{m+n+1} (m+n)!^2}{m! \,(m + 2 n+!)!}. $$