$\int {\cfrac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}dx}$ equals

Option b is correct

: Let $I = \int {\cfrac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {{e^x}x} \right)}}dx}$

Put $x{e^x} = t$ $\Rightarrow$ $\left( {{e^x} \cdot 1 + {e^x}x} \right)dx = dt$

$\therefore$ $I = \int {\cfrac{{dt}}{{{{\cos }^2}t}} = \int {{{\sec }^2}t\,dt} = \tan \,t + C = tan\left( {x{e^x}} \right) + C}$

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