Answer
$$\eqalign{
& \left( {\text{a}} \right)7 \cr
& \left( {\text{b}} \right) - 5 \cr
& \left( {\text{c}} \right) - 3 \cr
& \left( {\text{d}} \right)6 \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\int_1^{ac} {\frac{1}{t}} dt \cr
& {\text{Integrate}} \cr
& \int_1^{ac} {\frac{1}{t}} dt = \left[ {\ln t} \right]_1^{ac} \cr
& \left[ {\ln t} \right]_1^{ac} = \ln ac - \ln 1 \cr
& \left[ {\ln t} \right]_1^{ac} = \ln a + \ln c - 0 \cr
& {\text{Where }}\ln a = 2{\text{ and }}\ln c = 5 \cr
& \left[ {\ln t} \right]_1^{ac} = 2 + 5 - 0 \cr
& \left[ {\ln t} \right]_1^{ac} = 7 \cr
& \cr
& \left( {\text{b}} \right)\int_1^{1/c} {\frac{1}{t}} dt \cr
& {\text{Integrate}} \cr
& \int_1^{1/c} {\frac{1}{t}} dt = \left[ {\ln t} \right]_1^{1/c} \cr
& \left[ {\ln t} \right]_1^{1/c} = \ln \frac{1}{c} - \ln 1 \cr
& \left[ {\ln t} \right]_1^{1/c} = \ln 1 - \ln c - \ln 1 \cr
& \left[ {\ln t} \right]_1^{1/c} = - \ln c \cr
& {\text{Where }}\ln c = 5 \cr
& \left[ {\ln t} \right]_1^{1/c} = - 5 \cr
& \cr
& \left( {\text{c}} \right)\int_1^{a/c} {\frac{1}{t}} dt \cr
& {\text{Integrate}} \cr
& \int_1^{1/c} {\frac{1}{t}} dt = \left[ {\ln t} \right]_1^{a/c} \cr
& \left[ {\ln t} \right]_1^{a/c} = \ln \frac{a}{c} - \ln 1 \cr
& \left[ {\ln t} \right]_1^{a/c} = \ln a - \ln c - \ln 1 \cr
& \left[ {\ln t} \right]_1^{a/c} = \ln a - \ln c \cr
& {\text{Where }}\ln a = 2{\text{ and }}\ln c = 5 \cr
& \left[ {\ln t} \right]_1^{a/c} = 2 - 5 \cr
& \left[ {\ln t} \right]_1^{a/c} = - 3 \cr
& \cr
& \left( {\text{d}} \right)\int_1^{{a^3}} {\frac{1}{t}} dt \cr
& {\text{Integrate}} \cr
& \int_1^{{a^3}} {\frac{1}{t}} dt = \left[ {\ln t} \right]_1^{{a^3}} \cr
& \left[ {\ln t} \right]_1^{{a^3}} = \ln {a^3} - \ln 1 \cr
& \left[ {\ln t} \right]_1^{{a^3}} = 3\ln a - 0 \cr
& {\text{Where }}\ln a = 2{\text{ }} \cr
& \left[ {\ln t} \right]_1^{{a^3}} = 3\left( 2 \right) \cr
& \left[ {\ln t} \right]_1^{{a^3}} = 6 \cr} $$