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The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.

The new integration bounds are $x=2$ and $x=3$. When $u=0$, $x = 1 + e^0 = 1 + 1 = 2$.

Conditions: The substitution is $x = 1 + e^u$.; The original bounds are $u=0$ and $u=\ln 2$.; The exponential function $e^u$ is evaluated at these points.