Integral of $$$e^{x} \ln\left(x\right)$$$
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Find $$$\int e^{x} \ln\left(x\right)\, dx$$$.
Solution
For the integral $$$\int{e^{x} \ln{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.
Let $$$\operatorname{u}=\ln{\left(x \right)}$$$ and $$$\operatorname{dv}=e^{x} dx$$$.
Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (steps can be seen »).
Thus,
$${\color{red}{\int{e^{x} \ln{\left(x \right)} d x}}}={\color{red}{\left(\ln{\left(x \right)} \cdot e^{x}-\int{e^{x} \cdot \frac{1}{x} d x}\right)}}={\color{red}{\left(e^{x} \ln{\left(x \right)} - \int{\frac{e^{x}}{x} d x}\right)}}$$
This integral (Exponential Integral) does not have a closed form:
$$e^{x} \ln{\left(x \right)} - {\color{red}{\int{\frac{e^{x}}{x} d x}}} = e^{x} \ln{\left(x \right)} - {\color{red}{\operatorname{Ei}{\left(x \right)}}}$$
Therefore,
$$\int{e^{x} \ln{\left(x \right)} d x} = e^{x} \ln{\left(x \right)} - \operatorname{Ei}{\left(x \right)}$$
Add the constant of integration:
$$\int{e^{x} \ln{\left(x \right)} d x} = e^{x} \ln{\left(x \right)} - \operatorname{Ei}{\left(x \right)}+C$$
Answer
$$$\int e^{x} \ln\left(x\right)\, dx = \left(e^{x} \ln\left(x\right) - \operatorname{Ei}{\left(x \right)}\right) + C$$$A