Make the substitution $1 + 2x = t$ so that $\text{d}x = \frac{\text{d}t}{2}$ so you get:
$$I = \int \frac{(t-1)}{2}\cdot \frac{e^{2\left(\frac{(t-1)}{2}\right)}}{t^2}\frac{\text{d}t}{2}$$
and arranging the terms you easily get:
$$\frac{e^{-1}}{4}\int \frac{e^t}{t} - \frac{e^{t}}{t^2}\ \text{d}t$$
Now the first integral is a Special Function called the Exponential Integral function:
$$\int\frac{e^t}{t}\ \text{d}t = \text{Ei}(t)$$
and the second one can be performed by parts, giving the quite same result:
$$\int \frac{e^t}{t^2}\ \text{d}t = -\frac{e^t}{t} + \text{Ei}(t)$$
Putting together and you see the two Special Functions are cancelled by the minus sign, obtaining in the end the result of the integration in $\text{d}t$:
$$\frac{e^{-1}}{4}\frac{e^{t}}{t} \equiv \frac{e^{t-1}}{4t}$$
coming back to $x$:
$$I = \frac{e^{2x}}{4\cdot(1 + 2x)}$$
More about Exponential Integral
https://en.wikipedia.org/wiki/Exponential_integral
Final Remark
Don't forget about the various $C$ constants you can obtain from each integration, and you can set up them as zero!