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