Question
Evaluate the following integrals:
$\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$

Answer

Let $\text{I}=\frac{\text{x}+1}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)(1+\text{xe}^\text{x}-\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{\int(\text{x}+1)(1+\text{xe}^\text{x})}{\text(1+\text{xe}^\text{x})}\ \text{dx}-\int\frac{(\text{x}+1)(\text{xe}^\text{x})}{\text{x}(1+\text{xe}^\text{x})}\ \text{dx}$
$=\int\frac{(\text{x}+1)}{\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\int\frac{(\text{x}+1)\text{e}^\text{x}}{\text{xe}^\text{x}}\ \text{dx}-\int\frac{\text{e}^\text{x}(\text{x}+1)}{1+\text{xe}^\text{x}}\ \text{dx}$
$=\log|\text{xe}^\text{x}|-\log|1+\text{xe}^\text{x}|+\text{C}$
$\therefore\text{I}=\log\Big|\frac{\text{xe}^\text{x}}{1+\text{xe}^\text{x}}\Big|+\text{C}$

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