MCQ
$\int_{}^{} {{e^x}{{\sec }^2}({e^x})\;dx} $ =
- ✓$\tan ({e^x}) + k$
- B$\tan ({e^x})\;.\;e + k$
- C${e^x}\tan x + k$
- D$\frac{{\tan ({e^x})}}{{{e^x}}} + k$
Put ${e^x} = t \Rightarrow {e^x}dx = dt$
$\therefore \,\,\,I = \int_{}^{} {{{\sec }^2}t\,dt = \tan t + k = \tan ({e^x}) + k} $.
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