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For example: find $f$ if $f'(x) = \dfrac{e^{2x} + 4e^{-x}}{e^x}$ and $f(\ln 2) = 2$.

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    Its the same as finding an anti derivative and adding the appropriate constant to satisfy the condition $f(\ln 2)=2$ – 2017-01-07
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    It's the same as finding one particular antiderivative, and from that you can find an integral, as Jacky Chong's answer indicates. – 2017-01-07
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    @MichaelHardy of-topic question: why is \dfrac preferred over \frac? – 2017-01-07
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    @user2520938 : The "d" stands for "display". With "\frac a b" in an inline setting, one sees $\frac a b$ and in a displayed setting one sees: $$\frac a b$$ With \dfrac a b in an inline setting one sees $\dfrac a b$ rather than $\frac a b.$ In a displayed setting, the results are identical and I just write \frac. In this case, when the fraction is typographically moderately complicated expression, I think making it bigger helps legibility. – 2017-01-07

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Put $ t=e^{-x} , dt=-tdx$. the integral becomes

$$-\int (t^{-2}+4t)dt=\frac{1}{t}-2t^2+C$$

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

$$f(\ln(2))=2-\frac{1}{2}+C=2$$ $\implies C=\frac{1}{2}$.

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    Did you think of purpose of question? – 2017-01-08
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Hint: Using Fundamental Theorem of Calculus, we see that \begin{align} f(x) = f(\ln 2)+\int^x_{\ln 2} f'(t)\ dt =2+\int^x_{\ln 2}e^t+4e^{-2t}\ dt. \end{align}