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प्रश्न
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]
योग
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उत्तर
\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx\]
\[Let, 1 + \sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2dt\]
\[Now, \int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}}dx \]
\[ = 2\int t^2 dt\]
\[ = \frac{2}{3} t^3 + C\]
\[ = \frac{2}{3} \left( 1 + \sqrt{x} \right)^3 + C\]
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