Answer
$\displaystyle\int_0^1{\frac{1+12t}{1+3t}dt}=4-\ln 4$
Work Step by Step
$
I=\displaystyle\int_0^1{\frac{1+12t}{1+3t}dt}\\
I=\displaystyle\int_0^1{4-\frac{3}{1+3t}dt}\\
I=\displaystyle\int_0^1{4\ dt}-\int_0^1{\frac{3}{1+3t}dt}\\
I=\left[ 4t \right] _{0}^{1}-\left[ \ln \left| 1+3t \right| \right] _{0}^{1}\\
I=4-\left( \ln 4-\ln 1 \right) \\
I=4-\ln 4
$