Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 548: 25

$\displaystyle\int_0^1{\frac{1+12t}{1+3t}dt}=4-\ln 4$

$ 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 $