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प्रश्न
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
योग
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उत्तर
\[\int\frac{dx}{\sqrt{1 + 4 x^2}}\]
\[ = \int\frac{dx}{\sqrt{1 + \left( 2x \right)^2}}\]
\[\text{let 2x }= t\]
\[ \Rightarrow 2dx = dt\]
\[ \Rightarrow dx = \frac{dt}{2}\]
\[Now, \int\frac{dx}{\sqrt{1 + \left( 2x \right)^2}}\]
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{1 + t^2}} \]
\[ = \frac{1}{2} \text{ log} \left| t + \sqrt{1 + t^2} \right| + C \left[ \because \int\frac{dx}{\sqrt{x^2 + a^2}} = \text{ log} \left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \frac{1}{2} \text{ log }\left| 2x + \sqrt{1 + 4 x^2} \right| + C\]
\[ = \int\frac{dx}{\sqrt{1 + \left( 2x \right)^2}}\]
\[\text{let 2x }= t\]
\[ \Rightarrow 2dx = dt\]
\[ \Rightarrow dx = \frac{dt}{2}\]
\[Now, \int\frac{dx}{\sqrt{1 + \left( 2x \right)^2}}\]
\[ = \frac{1}{2}\int\frac{dt}{\sqrt{1 + t^2}} \]
\[ = \frac{1}{2} \text{ log} \left| t + \sqrt{1 + t^2} \right| + C \left[ \because \int\frac{dx}{\sqrt{x^2 + a^2}} = \text{ log} \left| x + \sqrt{x^2 + a^2} \right| + C \right]\]
\[ = \frac{1}{2} \text{ log }\left| 2x + \sqrt{1 + 4 x^2} \right| + C\]
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