Obtaining recurrence relation from integrals

Question

Consider, for $n\in\mathbb{Z}^+$, the integrals

$I(n)=\int_{0}^{1}x^n\cdot e^\frac{x}{2}dx$

They satisfy the recurrence relation

$I(n)=2\cdot\sqrt{e}-2\cdot n\cdot I(n-1)$ with $I(0)=2\cdot\sqrt{e}-2$

How would one start with the integrals $I(n)$ above to obtain the given recurrence relation from scratch?

I've seen examples of obtaining recurrence relations from integrals of the form
$\int_{a}^{b}\frac{x^n}{x+y}$
where $a,b,y\in\mathbb{R}$ are constants with $a<b$

Here, we would start by doing
$\int_{a}^{b}\frac{x^n}{x+y}=\int_{a}^{b}\frac{x^{n-1}(x+y-y)}{x+y}dx=...$

But how do we deal with other integrals not of this form?

Answer 1

By integrating by parts, one has $$ I(n)=\int_{0}^{1}x^n\cdot e^\frac{x}{2}dx=\left[ x^n\cdot 2e^\frac{x}{2}\right]_0^1-2n\int_{0}^{1}x^{n-1}\cdot e^\frac{x}{2}dx $$ that is

$$ I(n)=2\sqrt{e}-2n\cdot I(n-1),\quad n\ge1, $$

as announced.