Question

${\displaystyle\int}\dfrac{2020\left(x^{\mathrm{e}^{\pi}}-\mathrm{e}^{-{\gamma}}-1\right)}{\ln\left(x\right)}\,\mathrm{d}x$​

Answer 1

${\displaystyle \tt \red{{\int}\dfrac{2020\left(x^{\mathrm{e}^{\pi}}-\mathrm{e}^{-{\gamma}}-1\right)}{\ln\left(x\right)}\,\mathrm{d}x}}$

${ \red{= }\displaystyle\tt \color{red} { \int}\left(\dfrac{2020x^{\mathrm{e}^{\pi}}}{\ln\left(x\right)}-\dfrac{2020\mathrm{e}^{-{\gamma}}}{\ln\left(x\right)}-\dfrac{2020}{\ln\left(x\right)}\right)\mathrm{d}x}$

$\tt \red{{ = \displaystyle\int}\dfrac{x^{\mathrm{e}^{\pi}}}{\ln\left(x\right)}\,\mathrm{d}x+{}{{}{2020\mathrm{e}^{-{\gamma}}\left(-\mathrm{e}^{\gamma}-1\right)}}{\cdot}{\displaystyle\int}\dfrac{1}{\ln\left(x\right)}\,\mathrm{d}x}$

$\tt \color{red}{\displaystyle\int}\dfrac{x^{\mathrm{e}^{\pi}}}{\ln\left(x\right)}\,\mathrm{d}x$

$\tt \red{={\displaystyle\int}\dfrac{\mathrm{e}^{\left(\mathrm{e}^{\pi}+1\right)u}}{u}\,\mathrm{d}u}$

$\tt \red{={\displaystyle\int}\dfrac{\mathrm{e}^v}{v}\,\mathrm{d}v}$

$\tt \red{= Ei(v)}$

$\tt \red{= Ei(( {e}^{\pi} + 1)u)}$

$\tt \red{= Ei(( {e}^{\pi} + 1) \ln(x) )}$

$\tt \red{{\displaystyle\int}\dfrac{1}{\ln\left(x\right)}\,\mathrm{d}x}$

$\rm\red{ = li(x)}$

$\tt \red{{{2020}}{\displaystyle\int}\dfrac{x^{\mathrm{e}^{\pi}}}{\ln\left(x\right)}\,\mathrm{d}x+{}{{}{2020\mathrm{e}^{-{\gamma}}\left(-\mathrm{e}^{\gamma}-1\right)}}{\cdot}{\displaystyle\int}\dfrac{1}{\ln\left(x\right)}\,\mathrm{d}x}$

$\tt \red{=2020\operatorname{Ei}\left(\left(\mathrm{e}^{\pi}+1\right)\ln\left(x\right)\right)+2020\mathrm{e}^{-{\gamma}}\left(-\mathrm{e}^{\gamma}-1\right)\operatorname{li}\left(x\right)}$

${ \tt { \color{red}{\displaystyle\int}\dfrac{2020\left(x^{\mathrm{e}^{\pi}}-\mathrm{e}^{-{\gamma}}-1\right)}{\ln\left(x\right)}\,\mathrm{d}x}}$

$\tt \red{=2020\operatorname{Ei}\left(\left(\mathrm{e}^{\pi}+1\right)\ln\left(x\right)\right)+2020\mathrm{e}^{-{\gamma}}\left(-\mathrm{e}^{\gamma}-1\right)\operatorname{li}\left(x\right)+C}$

$\tt \red{=2020\operatorname{Ei}\left(\left(\mathrm{e}^{\pi}+1\right)\ln\left(x\right)\right)-2020\mathrm{e}^{-{\gamma}}\operatorname{li}\left(x\right)-2020\operatorname{li}\left(x\right)+C}$