Answer
$$1 - \ln 9$$
Work Step by Step
$$\eqalign{
& \int_{ - 1/2}^{1/2} {\frac{{{u^2} + 1}}{{{u^2} - 1}}} du \cr
& {\text{By the long division }}\frac{{{u^2} + 1}}{{{u^2} - 1}} = 1 + \frac{2}{{{u^2} - 1}} \cr
& \int_{ - 1/2}^{1/2} {\frac{{{u^2} + 1}}{{{u^2} - 1}}} du = \int_{ - 1/2}^{1/2} {\left( {1 + \frac{2}{{{u^2} - 1}}} \right)} du \cr
& = \int_{ - 1/2}^{1/2} {du} + 2\int_{ - 1/2}^{1/2} {\frac{1}{{{u^2} - 1}}du} \cr
& {\text{Decompose }}\frac{1}{{{u^2} - 1}}{\text{ into partial fractions}} \cr
& \frac{1}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{A}{{u + 1}} + \frac{B}{{u - 1}} \cr
& 1 = A\left( {u - 1} \right) + B\left( {u + 1} \right) \cr
& {\text{If }}u = - 1,\,\,\, \to A = - \frac{1}{2} \cr
& {\text{If }}u = 1,\,\,\,\,\,\, \to B = \frac{1}{2} \cr
& \frac{1}{{\left( {u + 1} \right)\left( {u - 1} \right)}} = \frac{{ - 1/2}}{{u + 1}} + \frac{{1/2}}{{u - 1}} = \frac{1}{2}\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right) \cr
& {\text{,then}} \cr
& \cr
& = \int_{ - 1/2}^{1/2} {du} + 2\left( {\int_{ - 1/2}^{1/2} {\frac{1}{2}\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right)du} } \right) \cr
& = \int_{ - 1/2}^{1/2} {du} + \int_{ - 1/2}^{1/2} {\left( {\frac{1}{{u - 1}} - \frac{1}{{u + 1}}} \right)du} \cr
& {\text{Integrate}} \cr
& = \left[ u \right]_{ - 1/2}^{1/2} + \left[ {\ln \left| {u - 1} \right| - \ln \left| {u + 1} \right|} \right]_{ - 1/2}^{1/2} \cr
& = \left[ u \right]_{ - 1/2}^{1/2} + \left[ {\ln \left| {\frac{{u - 1}}{{u + 1}}} \right|} \right]_{ - 1/2}^{1/2} \cr
& = \frac{1}{2} - \left( { - \frac{1}{2}} \right) + \ln \left| {\frac{{1/2 - 1}}{{1/2 + 1}}} \right| - \ln \left| {\frac{{ - 1/2 - 1}}{{ - 1/2 + 1}}} \right| \cr
& {\text{simplifying}} \cr
& = 1 + \ln \left| { - \frac{1}{3}} \right| - \ln \left| { - 3} \right| \cr
& = 1 + \ln \left( {\frac{1}{3}} \right) - \ln 3 \cr
& = 1 + \ln \left( {\frac{1}{9}} \right) \cr
& = 1 - \ln 9 \cr} $$
