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प्रश्न
उत्तर
\[\text{Let I} = \int\frac{1}{\sqrt{x}\left( \sqrt{x} + 1 \right)}dx\]
\[\text{Putting}\ \sqrt{x} + 1 = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{x}}dx = 2dt\]
\[ \therefore I = 2\int\frac{1}{t}dt\]
\[ =\text{ 2 }\text{ln}\left| t \right| + C\]
\[ = \text{2 }\text{ln} \left| \sqrt{x} + 1 \right| + C \left[ \because t = \sqrt{x} + 1 \right]\]
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