Question
Evaluate the following integrals:
$\int\text{e}^{\text{x}}(\log\text{x}+\frac{1}{\text{x}^2})\text{dx}$

Answer

We have,
$\text{I}=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x} ^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}+\frac{1}{\text{x}}-\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\int\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)-\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}+\int\text{e}^{\text{x}}\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{dx}$
$=\text{e}^{\text{x}}\Big(\log\text{x}-\frac{1}{\text{x}}\Big)+\text{C}$

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