$\int{ \frac{ 1 }{ 2 } \times \ln\left({t}\right) } \mathrm{d} t$
Use the property of integral $\begin{array} { l }\int{ a \times f\left( x \right) } \mathrm{d} x=a \times \int{ f\left( x \right) } \mathrm{d} x,& a \in ℝ\end{array}$$\frac{ 1 }{ 2 } \times \int{ \ln\left({t}\right) } \mathrm{d} t$
To use the partial integration formula, expand the expression into $\ln\left({t}\right) \times 1$$\frac{ 1 }{ 2 } \times \int{ \ln\left({t}\right) \times 1 } \mathrm{d} t$
Evaluate the integral using the partial integration formula$\frac{ 1 }{ 2 } \times \left( \ln\left({t}\right) \times t-\int{ t \times \frac{ 1 }{ t } } \mathrm{d} t \right)$
Cancel out the common factor $t$$\frac{ 1 }{ 2 } \times \left( \ln\left({t}\right) \times t-\int{ 1 } \mathrm{d} t \right)$
Use $\int{ 1 } \mathrm{d} x=x$ to evaluate the integral$\frac{ 1 }{ 2 } \times \left( \ln\left({t}\right) \times t-t \right)$
Substitute back $t=2x+1$$\frac{ 1 }{ 2 } \times \left( \ln\left({2x+1}\right) \times \left( 2x+1 \right)-\left( 2x+1 \right) \right)$
Simplify the expression$\frac{ 1 }{ 2 } \times \ln\left({2x+1}\right) \times \left( 2x+1 \right)-x-\frac{ 1 }{ 2 }$
Add the constant of integration $C \in ℝ$$\begin{array} { l }\frac{ 1 }{ 2 } \times \ln\left({2x+1}\right) \times \left( 2x+1 \right)-x+C,& C \in ℝ\end{array}$