a) \(\int\limits_1^2 {\frac{{x - 1}}{{{x^2}}}} dx = \int\limits_1^2 {\frac{1}{x}dx} - \int\limits_1^2 {\frac{1}{{{x^2}}}dx} = \int\limits_1^2 {\frac{1}{x}dx} - \int\limits_1^2 {{x^{ - 2}}dx} = \left. {\left( {\ln \left| x \right|} \right)} \right|_1^2 - \left. {\left( {\frac{{{x^{ - 1}}}}{{ - 1}}} \right)} \right|_1^2\)
\( = \left( {\ln 2 - \ln 1} \right) - \left( {\frac{{{2^{ - 1}}}}{{ - 1}} - \frac{{{1^{ - 1}}}}{1}} \right) = \ln 2 + \frac{3}{2}\)
b) \(\int\limits_0^\pi {\left( {1 + 2{{\sin }^2}\frac{x}{2}} \right)dx} = \int\limits_0^\pi {\left( {1 + 1 - \cos x} \right)dx} = \int\limits_0^\pi {\left( {2 - \cos x} \right)dx} = 2\int\limits_0^\pi {dx} - \int\limits_0^\pi {\cos xdx} \)
\( = 2\left. {\left( x \right)} \right|_0^\pi - \left. {\left( {\sin x} \right)} \right|_0^\pi = 2\left( {\pi - 0} \right) - \left( {\sin \pi - \sin 0} \right) = 2\pi \)
c) \(\int\limits_{ - 2}^1 {{{\left( {x - 2} \right)}^2}dx} + \int\limits_{ - 2}^1 {\left( {4x - {x^2}} \right)dx} = \int\limits_{ - 2}^1 {\left[ {{{\left( {x - 2} \right)}^2} + 4x - {x^2}} \right]dx = \int\limits_{ - 2}^1 {\left( {{x^2} - 4x + 4 + 4x - {x^2}} \right)dx} } \)
\( = \int\limits_{ - 2}^1 {4dx} = \left. {4x} \right|_{ - 2}^1 = 4.1 - 4\left( { - 2} \right) = 12\)