\(\text{Consider the given integral}\)

\(\int_0^3\frac{2x-1}{x+1}\)

\(\text{Let }x+1=u\Rightarrow dx=du\)

\(\text{Also, when }x=0\Rightarrow u=1\text{ when }x=3\Rightarrow u=4\)

\(\text{So, the integral becomes:}\)

\(\int_1^4\frac{2(u-1)-1}{u}du\)

\(\Rightarrow\int_1^4\frac{2u-3}{u}du\)

\(\Rightarrow\int_1^4\frac{2u}{u}-\frac{3}{u}du\)

\(\Rightarrow\int_1^4 2-\frac{3}{u}du\)

\(=2u-3\ln (u)|_1^4\)

\(=2(4)-3\ln(4)-2(1)+3\ln(1)\)

\(\Rightarrow8-3\ln(4)-2+3(0)\)

\(\Rightarrow6-3\ln(2^2)\)

\(\Rightarrow6-6\ln(2)\)

\(\text{Hence, }\int_0^3\frac{2x-1}{x+1}dx=6-6\ln(2)\)

\(\int_0^3\frac{2x-1}{x+1}\)

\(\text{Let }x+1=u\Rightarrow dx=du\)

\(\text{Also, when }x=0\Rightarrow u=1\text{ when }x=3\Rightarrow u=4\)

\(\text{So, the integral becomes:}\)

\(\int_1^4\frac{2(u-1)-1}{u}du\)

\(\Rightarrow\int_1^4\frac{2u-3}{u}du\)

\(\Rightarrow\int_1^4\frac{2u}{u}-\frac{3}{u}du\)

\(\Rightarrow\int_1^4 2-\frac{3}{u}du\)

\(=2u-3\ln (u)|_1^4\)

\(=2(4)-3\ln(4)-2(1)+3\ln(1)\)

\(\Rightarrow8-3\ln(4)-2+3(0)\)

\(\Rightarrow6-3\ln(2^2)\)

\(\Rightarrow6-6\ln(2)\)

\(\text{Hence, }\int_0^3\frac{2x-1}{x+1}dx=6-6\ln(2)\)