Answer
$f(x)=\frac{(2x-1)e^{2x}}{1-e^{-x}}$
Work Step by Step
$\int_1^xf(t)dt=(x-1)e^{2x}+\int_1^xe^{-t}f(t)dt$ (Differentiate both sides with respect to $x$)
$\frac{d}{dx}(\int_1^xf(t)dt)=\frac{d}{dx}((x-1)e^{2x}+\int_1^xe^{-t}f(t)dt)$ (Apply the Sum Rule for Derivatives)
$\frac{d}{dx}\int_1^xf(t)dt=\frac{d}{dx}((x-1)e^{2x})+\frac{d}{dx}\int_1^xe^{-t}f(t)dt$ (Apply the Product Rule for Derivatives)
$\frac{d}{dx}\int_1^xf(t)dt=\frac{d}{dx}(x-1)\cdot e^{2x}+(x-1)\frac{d}{dx}e^{2x}+\frac{d}{dx}\int_1^xe^{-t}f(t)dt$ (Apply the Fundamental Theorem of Calculus in Part 1)
$f(x)=1\cdot e^{2x}+(x-1)\cdot 2e^{2x}+e^{-x}f(x)$ (Find $f(x)$)
$f(x)=1e^{2x}+(2x-2)e^{2x}+e^{-x}f(x)$
$f(x)=(2x-1)e^{2x}+e^{-x}f(x)$
$f(x)-e^{-x}f(x)=(2x-1)e^{2x}$
$(1-e^{-x})f(x)=(2x-1)e^{2x}$
$f(x)=\frac{(2x-1)e^{2x}}{1-e^{-x}}$
Thus,
$f(x)=\frac{(2x-1)e^{2x}}{1-e^{-x}}$
