Answer
$$\frac{{2\sinh bx}}{x}$$
Work Step by Step
$$\eqalign{
& \int_{ - b}^b {{e^{xt}}} dt \cr
& {\text{Integrate with respect to }}t \cr
& \int_{ - b}^b {{e^{xt}}} dt = \frac{1}{x}\left[ {{e^{xt}}} \right]_{ - b}^b \cr
& {\text{Evaluate}} \cr
& \frac{1}{x}\left[ {{e^{xt}}} \right]_{ - b}^b = \frac{1}{x}\left[ {{e^{bx}} - {e^{ - bt}}} \right] \cr
& = \frac{{{e^{bx}} - {e^{ - bx}}}}{x} \cr
& {\text{Let }}u = bx,{\text{ }}x = \frac{u}{b} \cr
& \frac{{{e^{bx}} - {e^{ - bx}}}}{x} = \frac{{{e^u} - {e^{ - u}}}}{{u/b}} \cr
& = \frac{b}{u}\left( {{e^u} - {e^{ - u}}} \right) \cr
& {\text{Multiply and divide by 2}} \cr
& = \frac{{2b}}{u}\left( {\frac{{{e^u} - {e^{ - u}}}}{2}} \right) \cr
& {\text{By the definition of the hyperbolic sine}} \cr
& = \frac{{2b}}{u}\sinh u \cr
& {\text{Write in terms of }}bx \cr
& = \frac{{2b}}{{bx}}\sinh bx \cr
& = \frac{{2\sinh bx}}{x} \cr} $$
