Answer
$$ x = \frac{6t}{t+2}-1, \quad (t\gt 0)$$
Work Step by Step
$$\eqalign{
& \left( {{t^2} + 2t} \right)\frac{{dx}}{{dt}} = 2x + 2 \cr
& {\text{Separate the variables}} \cr
& \frac{1}{{2x + 2}}dx = \frac{1}{{{t^2} + 2t}}dt \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\frac{1}{{2x + 2}}} dx = \int {\frac{1}{{{t^2} + 2t}}} dt \cr
& \cr
& {\text{Decompose }}\frac{1}{{{t^2} + 2t}}{\text{ into partial fractions}} \cr
& \frac{1}{{{t^2} + 2t}} = \frac{1}{{t\left( {t + 2} \right)}} \cr
& \frac{1}{{t\left( {t + 2} \right)}} = \frac{A}{t} + \frac{B}{{t + 2}} \cr
& {\text{Multiply by }}t\left( {t + 2} \right){\text{ and simplify}} \cr
& 1 = A\left( {t + 2} \right) + Bt \cr
& {\text{if }}t = 0,\,\,\,A = 1/2 \cr
& {\text{if }}t = - 2,\,\,\,B = - 1/2 \cr
& \frac{1}{{t\left( {t + 2} \right)}} = \frac{{1/2}}{t} - \frac{{1/2}}{{t + 2}} \cr
& \cr
& \int {\frac{1}{{2x + 2}}} dx = \int {\frac{1}{{{t^2} + 2t}}} dt \cr
& \int {\frac{1}{{2x + 2}}} dx = \int {\left( {\frac{{1/2}}{t} - \frac{{1/2}}{{t + 2}}} \right)} dt \cr
& {\text{Integrate}} \cr
& \frac{1}{2}\ln \left| {2x + 2} \right| = \frac{1}{2}\ln \left| t \right| - \frac{1}{2}\ln \left| {t + 2} \right| + C \cr
& \cr
& {\text{Use }}x\left( 1 \right) = 1 \cr
& \frac{1}{2}\ln \left| {2 + 2} \right| = \frac{1}{2}\ln \left| 1 \right| - \frac{1}{2}\ln \left| {1 + 2} \right| + C \cr
& \frac{1}{2}\ln 4 = - \frac{1}{2}\ln \left| 3 \right| + C \cr
& \frac{1}{2}\ln 4 + \frac{1}{2}\ln \left| 3 \right| = + C \cr
& C = \frac{1}{2}\ln \left( {4 \cdot 3} \right) \cr
& C = \frac{1}{2}\ln 12 \cr
& \cr
& \frac{1}{2}\ln \left| {2x + 2} \right| = \frac{1}{2}\ln \left| t \right| - \frac{1}{2}\ln \left| {t + 2} \right| + \frac{1}{2}\ln 12 \cr
& {\text{Multiply by 2}} \cr
& \ln \left| {2x + 2} \right| = \ln \left| t \right| - \ln \left| {t + 2} \right| + \ln 12 \cr
& {\text{Take the exponent of both sides}} \cr
& e^{\ln \left| {2x + 2} \right|} = e^{\ln \left| \frac{t}{t+2} \right| + \ln 12 } \cr
& 2x + 2 = \frac{12t}{t+2}\cr
& x = \frac{6t}{t+2}-1\cr}
$$
