Calculate: integral of ln(2x+1) x

Expression: $\int{ \ln\left({2x+1}\right) } \mathrm{d} x$

Use the substitution $t=2x+1$ to transform the integral

$\int{ \frac{ 1 }{ 2 } \times \ln\left({t}\right) } \mathrm{d} t$

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 $C \in ℝ$

$\begin{array} { l }\frac{ 1 }{ 2 } \times \ln\left({2x+1}\right) \times \left( 2x+1 \right)-x+C,& C \in ℝ\end{array}$

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