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The two functions you mention actually differ only by a constant. Note that $\ln(5x)/5+C=\ln(x)/5+(\ln(5)/5+C)=\ln(x)/5+D$. They both have the same derivative and are both anti derivatives of $1/5x$. If you choose initial conditions, say demanding that the anti derivative satisfies $F(1/5)=0$, and you work out the correct constant, then you will get the ...


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As for your first example, $$ \ln(5x) = \ln 5 + \ln x$$ so $$ \frac{\ln(5x)}5 +C = \frac{\ln x+\ln 5}5 +C = \frac{\ln x}5+ \color{green}{\frac{\ln 5}5 +C}$$ and the green part becomes a new constant: $$ \frac{\ln(5x)}5 +C = \frac{\ln x}5+ \color{green}{C}$$ The indefinite integral is not a single function, but rather a family of functions differing by (any) ...


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All integrands of a function differ by a constant, i.e. if you have $f_1,f_2$ such that $f_1' = f_2' = f$, then $f_1=f_2 + C$ for some constant $C$. In your examples: The first example: $$f_1(x) = \frac15 \ln(5x) = \frac{1}{5}(\ln 5 + \ln x) = \frac15\ln 5 + \frac15\ln x=\frac{1}{5}\ln 5 + f_2(x),$$ meaning that the two functions differ only by a ...


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$$\lim _{h\to 0}\left(\frac{\sqrt{7\left(a+h\right)}-\sqrt{7a}}{h}\right)$$ Apply L'Hopital's Rule: $$\frac{d}{dh}\left(\sqrt{7\left(a+h\right)}-\sqrt{7a}\right)=\frac{\sqrt{7}}{2\sqrt{a+h}}$$ $$\frac{d}{dh}\left(h\right)=1$$ Then $$\lim _{h\to 0}\left(\frac{\frac{\sqrt{7}}{2\sqrt{a+h}}}{1}\right) = \lim _{h\to 0}\left(\frac{\sqrt{7}}{2\sqrt{a+h}}\right)= ...


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Hint: You may always multiply by one or add zero and it will not change anything. Here, let us multiply by $1$ in the form $$\frac{\sqrt{7(a+h)}+\sqrt{7a}}{\sqrt{7(a+h)}+\sqrt{7a}}$$ What happens?


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It is not. The moment generating function for a random variable $X$ is defined as $$MGF_X(t) = E\left[e^{Xt}\right]$$ where $t$ is the argument of the function, a real number. If $X$ is a degenerate random variable, a constant, $X =c$, then plugging this into the above we have $$MGF_c(t) = E\left[e^{ct}\right] = e^{ct}$$ since the function $e^{ct}$ is ...



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