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Question
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
Sum
Solution
\[\int\left( \frac{x + 1}{x} \right) \cdot \left( x + \log x \right)^2 dx\]
\[\text{Let x} + \log x = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x} \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( \frac{x + 1}{x} \right) dx = dt\]
\[Now, \int\left( \frac{x + 1}{x} \right) \cdot \left( x + \log x \right)^2 dx\]
\[ = \int t^2 dt\]
\[ = \frac{t^3}{3} + C\]
\[ = \frac{\left( x + \log x \right)^3}{3} + C\]
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