Answer
$$\frac{{{e^8} - {e^4}}}{4} - \frac{1}{6}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\left( {{e^{4u}} - \frac{1}{{{{\left( {u + 1} \right)}^2}}}} \right)} du \cr
& {\text{using the property of the exponents }}\frac{1}{{{z^n}}} = {z^{ - n}} \cr
& \int_1^2 {\left( {{e^{4u}} - {{\left( {u + 1} \right)}^{ - 2}}} \right)} du \cr
& {\text{integrate by using }}\int_a^b {{e^{kt}}} dt = \left( {\frac{{{e^{kt}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {{t^n}} dt = \left[ {\frac{{{t^{n + 1}}}}{{n + 1}}} \right]_a^b{\text{ }} \cr
& = \left[ {\frac{{{e^{4u}}}}{4} - \frac{{{{\left( {u + 1} \right)}^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_1^2 \cr
& = \left[ {\frac{{{e^{4u}}}}{4} - \frac{{{{\left( {u + 1} \right)}^{ - 1}}}}{{ - 1}}} \right]_1^2 \cr
& = \left[ {\frac{{{e^{4u}}}}{4} + \frac{1}{{u + 1}}} \right]_1^2 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{{{e^{4\left( 2 \right)}}}}{4} + \frac{1}{{2 + 1}}} \right) - \left( {\frac{{{e^{4\left( 1 \right)}}}}{4} + \frac{1}{{1 + 1}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{{{e^8}}}{4} + \frac{1}{3} - \frac{{{e^4}}}{4} - \frac{1}{2} \cr
& = \frac{{{e^8}}}{4} - \frac{{{e^4}}}{4} - \frac{1}{6} \cr
& = \frac{{{e^8} - {e^4}}}{4} - \frac{1}{6} \cr} $$
