Explanation

(b)
$\int_{b - 1}^{b}{\frac{e^{- t}}{t - b - 1}dt = \int_{- 1}^{0}{\frac{e^{- (t + b)}}{(t + b) - b - 1}dt}}$
$= e^{- b}\int_{- 1}^{0}{\frac{e^{- t}}{t - 1}dt}$
$= e^{- b}\int_{0}^{1}{\frac{e^{- (t - 1)}}{t - 2}dt}$
Put $t - 1 = - s$
$\Rightarrow \ dt = - ds$
$= - e^{- b}\int_{1}^{0}{\frac{e^{s}}{- (s + 1)}ds}$
$= e^{- b}\int_{1}^{0}{\frac{e^{s}}{s + 1}ds}$
$= - e^{- b}\int_{0}^{1}{\frac{e^{t}}{t + 1}dt}$
$= - ae^{- b}$