Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$

Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.

Following the Ishan Banerjee comment let $t=x^n$ hence $dx=\frac{1}{n}t^{\frac{1}{n}-1}dt$ and then $$ (2n+1) \int_0 ^{1} x^n e^x dx=\frac{2n+1}{n}\int_0^1t^{1/n}e^{t^{1/n}}dt\to2e$$ by using the dominated convergence theorem.

Let $I_n=\int\limits_0^1x^n\mathrm e^x\mathrm dx$. By integration by parts, $(n+1)I_n=\left.x^{n+1}\mathrm e^x\right|_0^1-I_{n+1}=\mathrm e-I_{n+1}$. Now, $0\leqslant I_{n+1}\leqslant I_n$ hence $(n+1)I_n\leqslant\mathrm e\leqslant(n+2)I_n$.

This is enough to show that $$ \left(2-\frac3n\right)\cdot\mathrm e\leqslant(2n+1)I_n\leqslant2\mathrm e, $$ hence $(2n+1)I_n\to2\mathrm e$.