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I differentiated both sides and got an answer $f(x)=\frac{e^{2x} + 2xe^{2x}}{1-e^{-x}}$. I think this might be correct, but I'm not sure if how I got there was correct. I just blindly differentiated without being certain of the rules and would appreciate if someone can provide an answer that explains the process in a little detail.

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  • $\begingroup$ Yes, differentiation is the way to go here; have you heard of the fundamental theorem of calculus? $\endgroup$ – Zain Patel Jul 13 '16 at 11:30
  • $\begingroup$ @ZainPatel Heard of it, yes, but I wouldn't say I've ever properly understood it. $\endgroup$ – math101 Jul 13 '16 at 11:31
  • $\begingroup$ It basically says that if you differentiate something like $\int_0^{x} f(t) \, \mathrm{d}t$, then you're differentiating $F(x) - F(0)$, but the $F(0)$ term disappears because it is a constant, and the derivative of $F(x)$ is $f(x)$. So $\frac{d}{dx} \int_0^x f(t) \, \mathrm{d}t = f(x)$ $\endgroup$ – Zain Patel Jul 13 '16 at 11:32
  • $\begingroup$ @ZainPatel I get it now. Thanks so much. $\endgroup$ – math101 Jul 13 '16 at 11:36
  • $\begingroup$ The integral on the right hand side isn't any different. Think of differentiating an integral as simply removing the integral sign, sort of like squaring a square root is just removing the square root symbol (as a very rough analogy) - so, if you take $g(t) = e^{-t} f(t)$ then $\frac{d}{dx}\int_0^x g(t)\, \mathrm{d}t = g(t) = e^{-t} f(t)$. $\endgroup$ – Zain Patel Jul 13 '16 at 11:36
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If we differentiate both sides of the equality, using the fundamental theorem of calculus, we get $$f(x) = 2xe^{2x} + e^{2x} + e^{-x}f(x)$$

re-arranging for $f(x)$, we get $$f(x) (1 - e^{-x}) = e^{2x}(1 + 2x) \Rightarrow f(x) = \frac{(1+2x)e^{2x}}{1 - e^{-x}}$$

as you got.


The fundamental theorem of calculus states that $\frac{d}{dx} \int_0^x f(t) \, \mathrm{d}t = f(x)$.

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  • $\begingroup$ But for what you have found, is the integral $\int_0^x f(t)dt$ convergent ? (for $x>0$, say) $\endgroup$ – Kelenner Jul 13 '16 at 13:35

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