Answer
$$x = \frac{{4{e^2}}}{{1 + {e^2}}}$$
Work Step by Step
$$\eqalign{
& \int_2^x {\frac{1}{{t\left( {4 - t} \right)}}} dt \cr
& {\text{Decompose the integrand into partial fractions}} \cr
& \frac{1}{{t\left( {4 - t} \right)}} = \frac{A}{t} + \frac{B}{{4 - t}} \cr
& 1 = A\left( {4 - t} \right) + Bt \cr
& {\text{Let }}t = 0 \Rightarrow A = \frac{1}{4} \cr
& {\text{Let }}t = 4 \Rightarrow B = \frac{1}{4} \cr
& \frac{1}{{t\left( {4 - t} \right)}} = \frac{1}{{4t}} + \frac{1}{{4\left( {4 - t} \right)}} \cr
& {\text{,then}} \cr
& \int_2^x {\frac{1}{{t\left( {4 - t} \right)}}} dt = \int_2^x {\left( {\frac{1}{{4t}} + \frac{1}{{4\left( {4 - t} \right)}}} \right)} dt \cr
& \int_2^x {\left( {\frac{1}{{4t}} + \frac{1}{{4\left( {4 - t} \right)}}} \right)} dt = \frac{1}{2} \cr
& {\text{Integrating}} \cr
& \left[ {\frac{1}{4}\ln t - \frac{1}{4}\ln \left| {4 - t} \right|} \right]_2^x = \frac{1}{2} \cr
& \left[ {\frac{1}{4}\ln \left| {\frac{t}{{4 - t}}} \right|} \right]_2^x = \frac{1}{2} \cr
& \frac{1}{4}\ln \left| {\frac{x}{{4 - x}}} \right| - \frac{1}{4}\ln \left| {\frac{2}{{4 - 2}}} \right| = \frac{1}{2} \cr
& {\text{Solve for }}x \cr
& \frac{1}{4}\ln \left| {\frac{x}{{4 - x}}} \right| - \frac{1}{4}\ln \left| 1 \right| = \frac{1}{2} \cr
& \frac{1}{4}\ln \left| {\frac{x}{{4 - x}}} \right| = \frac{1}{2} \cr
& \ln \left| {\frac{x}{{4 - x}}} \right| = 2 \cr
& \frac{x}{{4 - x}} = {e^2} \cr
& x = 4{e^2} - {e^2}x \cr
& x + {e^2}x = 4{e^2} \cr
& x = \frac{{4{e^2}}}{{1 + {e^2}}} \cr} $$
