A) \[\tan (x{{e}^{x}})+c\]
B) \[\sec (x{{e}^{x}})\tan (x{{e}^{x}})+c\]
C) \[-\tan (x{{e}^{x}})+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\frac{{{e}^{x}}(x+1)}{{{\cos }^{2}}(x{{e}^{x}})}=\int_{{}}^{{}}{{{e}^{x}}(x+1){{\sec }^{2}}(x{{e}^{x}})dx}}\] Putting \[x{{e}^{x}}=t\Rightarrow (x+1){{e}^{x}}dx=dt\], we get \[\int_{{}}^{{}}{{{\sec }^{2}}t\,dt=\tan t+c}=\tan (x{{e}^{x}})+c.\]
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