[tex]10^{\ln x}=e^{\ln 10^{\ln x}}=e^{\ln x\cdot\ln 10}=(e^{\ln x})^{\ln10}=x^{\ln10}[/tex]
So we have
[tex]\displaystyle\int10^{\ln x}\,\mathrm dx=\int x^{\ln10}\,\mathrm dx}=\dfrac{x^{1+\ln10}}{1+\ln10}+C[/tex]
which, as the above manipulation showed, is equivalent to
[tex]\dfrac{x10^{\ln x}}{1+\ln10}+C[/tex]
