This comes from integrating the ODE
The ODE to solve is
\begin{align*}
{\frac {\rm d}{{\rm d}t}}x \left( t \right) =a+b{{\rm e}^{gt}}+ \left( c+d{{\rm e}^{-gt}} \right) x \left( t \right)
\end{align*}
In canonical form, the ODE is written as
\begin{align*}
x' &= F(t,x)\\
&= {{\rm e}^{-gt}}dx+b{{\rm e}^{gt}}+cx+a
\end{align*}
The ODE is linear in $x$ and has the form
$$
x' = x f(t) + g(t)
$$
Where $f(t) = c+d{{\rm e}^{-gt}}$ and $g(t) = b{{\rm e}^{gt}}+a$.
Writing the ODE as
\begin{align*}
x' - \left( \left( c+d{{\rm e}^{-gt}} \right) x\right) &= b{{\rm e}^{gt}}+a\\
x' - \left( c+d{{\rm e}^{-gt}} \right) x &= b{{\rm e}^{gt}}+a
\end{align*}
Therefore the integrating factor $\mu$ is
$$
\mu = e^{\int -c-d{{\rm e}^{-gt}}\mathop{\mathrm{d}t}} = {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}
$$
The ode becomes
\begin{align*}
\frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}t}} \mu x &= \mu \left(b{{\rm e}^{gt}}+a\right) \\
\frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}t}} \left(x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) &= \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\\
\mathrm{d} \left(x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) &= \left( \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\right) \mathrm{d} t
\end{align*}
Integrating both sides gives
\begin{align*}
x{{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} &= \int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\,{\rm d}t + C_1
\end{align*}
Dividing both sides by the integrating factor $\mu={{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}$
results in
$$
x = {1\int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}}\,{\rm d}t \left( {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} \right) ^{-1}}+{{\it \_C1} \left( {{\rm e}^{-ct+{\frac {d{{\rm e}^{-gt}}}{g}}}} \right) ^{-1}}
$$
Simplifying the solution gives
$$
x={{\rm e}^{{\frac {ctg-d{{\rm e}^{-gt}}}{g}}}} \left( \int \! \left( b{{\rm e}^{gt}}+a \right) {{\rm e}^{-{\frac { \left( ctg{{\rm e}^{gt}}-d \right) {{\rm e}^{-gt}}}{g}}}}\,{\rm d}t+{\it \_C1} \right)
$$
Now, instead of writing $\int \!b{\mathrm{e}^{{\frac{-ctg+{g}^{2}t+d{\mathrm{%
e}^{-gt}}}{g}}}}+{\mathrm{e}^{{\frac{-ctg+d{\mathrm{e}^{-gt}}}{g}}}}a\,%
\mathrm{d}t$, Mathematica wrties as
\begin{equation*}
\int_{1}^{t}b{\mathrm{e}^{{\frac{-cKg+{g}^{2}K+d{\mathrm{e}^{-gK}}}{g}}}}+%
{\mathrm{e}^{{\frac{-cKg+d{\mathrm{e}^{-gK}}}{g}}}}a\,\mathrm{d}K
\end{equation*}
i.e. it uses $K[n]$ as dummy variable of integration inside the integral. That
is all. The choice of 1 as lower limit vs. 0 is arbitrary. Mathematica always uses 1 for lower limit in such cases.
1
inK[1]
just means it's the first integration dummy variable needed. If you had many you'd getK[1], K[2], ...
$\endgroup$C[1]
nicely formatted, they should do something similar with theK[1]
! $\endgroup$