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प्रश्न
\[\int\frac{\sec^2 x}{1 - \tan^2 x} dx\]
बेरीज
उत्तर
` ∫ sec^2 x /{1- tan^2 x }` dx
\[\text{let }\tan x = t\]
\[ \Rightarrow \sec^2 \text{ x dx }= dt\]
Now, ` ∫ sec^2 x /{1- tan^2 x }` dx
\[ = \int\frac{dt}{1 - t^2}\]
\[ = \frac{1}{2} \text{ log } \left| \frac{1 + t}{1 - t} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| \frac{1 + \tan x}{1 - \tan x} \right| + C\]
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