$\frac{\text{dy}}{\text{dx}}=\frac{\text{x e }^\text{x}\log\text{x}+\text{e}^\text{x}}{\text{x}\cos\text{y}}$
$\Rightarrow\text{x}\cos\text{y dy}=(\text{x e}^\text{x}\log\text{x}+\text{e}^\text{x})\ \text{dx}$
$\Rightarrow\cos\text{y dy}=\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\ \text{dx}$
Integrating both sides, we get
$\int\cos\text{y dy}=\int\Big(\text{e}^\text{x}\log\text{x}+\frac{1}{\text{x}}\text{e}^\text{x}\Big)\text{dx}$
$\Rightarrow\sin\text{y}=\log\text{x}\int\text{e} ^\text{x}\text{dx}-\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}+\int\frac{1}{\text{x}}\text{e}^\text{x}\text{dx}$
$\Rightarrow\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$
Hence, $\sin\text{y}=\text{e}^\text{x}\log\text{x}+\text{C}$ is the required solution.
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$\int\frac{\log\text{x}}{(\text{x}+1)^2}\text{dx}$