Integrate $\int{ e^{{x}^{2}-x} \cdot x \cdot e^x} dx$ 
Integrate $\int{ e^{{x}^{2}-x} \cdot x \cdot e^x} dx$

I'd like to know how to do it because I need it for another task. Here is what I tried:
$$\int{ e^{{x}^{2}-x} \cdot x \cdot e^x} dx = \int{e^{x^{2}-x+x} \cdot x} \text{ }dx = \int{e^{x^{2}} \cdot x} \text{ }dx$$
Now substitute (especially at this step I'm not sure). Let $s=x^2$, then:
$$s'=2x \Leftrightarrow 2x = \frac{ds}{dx} \iff dx = \frac{1}{2x} ds$$
Insert these into $\int{e^{x^{2}} \cdot x} \text{ }dx$: 
$$\int{e^{s} \cdot x} \cdot \frac{1}{2x} \text{ }ds= \frac{1}{2}\int{e^{s} ds} = \frac{1}{2}e^{s}+c = \frac{1}{2}e^{x^{2}}+c$$

I hope everything is alright? If it's correct, is there a faster way to solve it? Because substitution is confusing for me :s

You can always differentiate your answer and see what you get. In this case
$$
\frac{d}{dx}\left(\frac{1}{2}e^{x^{2}}+c\right)=\frac{1}{2}2x e^{x^{2}}=xe^{x^{2}}
$$
by the chain rule. So you are correct.
