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