Assuming no revenue from no calculators being sold, so that [tex]R(0)=0[/tex], we have
[tex]R(12)=R(0)+\displaystyle\int_0^{12}R'(x)\,\mathrm dx[/tex]
[tex]R(12)=\displaystyle\int_0^{12}(x+1)\ln(x+1)\,\mathrm dx[/tex]
Integrate by parts, taking
[tex]u=\ln(x+1)\implies\mathrm du=\dfrac{\mathrm dx}{x+1}[/tex]
[tex]\mathrm dv=x+1\,\mathrm dx\implies v=\dfrac{(x+1)^2}2[/tex]
Then
[tex]R(12)=\displaystyle\frac{(x+1)^2}2\ln(x+1)\bigg|_0^{12}-\frac12\int_0^{12}x+1\,\mathrm dx[/tex]
[tex]R(12)=\dfrac{169}2\ln13-\dfrac{(x+1)^2}4\bigg|_0^{12}[/tex]
[tex]R(12)=\dfrac{169}2\ln13-\left(\dfrac{169}4-\dfrac14\right)\approx174.74[/tex]
