Evaluate integral $\int e^{\mu (j-l)t}(1-e^{-\mu t})^{j-l-1}dt$

I have to evaluate the following integral, but I don't have any ideas how to do it.

$$\int e^{\mu (j-l)t}(1-e^{-\mu t})^{j-l-1}dt$$

Can anyone help me with this?

For simplicity, write $x=e^{\mu t}$ and $n=j-l$, the integrand is $x^n\left (1-x^{-1}\right )^{n-1}$. Using the Binomial Theorem we get$$\left (1-x^{-1}\right )^{n-1}=\sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^kx^{-k}$$then$$x^n\left (1-x^{-1}\right )^{n-1}=\sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^kx^{n-k}.$$Therefore$$e^{\mu nt}\left (1-e^{-\mu t}\right )^{n-1}=\sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^ke^{\mu t(n-k)}.$$Finally\begin{align*}\int e^{\mu nt}\left (1-e^{-\mu t}\right )^{n-1}\,dt & =\int \sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^ke^{\mu t(n-k)}\,dt \\ & =\sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^k\int e^{\mu t(n-k)}\,dt \\ & =\sum \limits _{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{e^{\mu t(n-k)}}{\mu (n-k)}+C \\ & =\sum \limits _{k=0}^{j-l-1}\binom{j-l-1}{k}(-1)^k\frac{e^{\mu t(j-l-k-1)}}{\mu (j-l-k)}+C. \end{align*}