Integration - finding an explicit formula The question in my textbook asks: > If $f$ is a continuous function such that $$\int\limits_0^x{f(t)dt}=xe^{2x}+\int\limits_0^x{e^{-t}f(t)dt}$$ for all $x$, find an explicit formula for $f(x)$. My working goes as follows: I decided to analyse the equation as an integration by parts $\left(\int udv=uv-\int vdu\right)$, so $uv]_0^x=xe^{2x}\\\ \therefore \text{a possible substitution is}\\\ \quad u=t,\qquad v=e^{2t}\\\ \quad du=dt,\quad dv=2e^{2t}dt$ and $\int\limits_0^xvdu=\int\limits_0^x{e^{-t}f(t)dt}\\\ \therefore e^{-t}f(t)=e^{2t}\\\ \quad f(t)=e^{3t}$ This looks sound until I try equating $\int\limits_0^xudv=\int\limits_0^x f(t)dt$, whereupon I get $f(t)=2te^{2t}$. I think I don't quite understand what an explicit formula is.

Notice that if you differentiate the equation:

$$f(x) = e^{2x} + 2xe^{2x} + e^{-x}*f(x)$$

$$f(x)(1 - e^{-x}) = e^{2x} + 2xe^{2x}$$

$$f(x) = \frac{e^{2x} + 2xe^{2x}}{1 - e^{-x}}$$