Answer
$$h'(x) = xe^x$$
Work Step by Step
$$h(x) = \int\limits^{e^x}_1 {ln(t)dt}$$
Using FTC 1, substitue in the upper bound for t and multiply by the derivative of the upper bound.
$$h'(x) = ln(e^x) \times (e^x)'$$
Simplify. [Note that $ln(e^x) = x$]
$$h'(x) = xe^x$$