A) \[2\log \cos (x{{e}^{x}})+c\]
B) \[\sec (x{{e}^{x}})+c\]
C) \[\tan (x{{e}^{x}})+c\]
D) \[\tan (x+{{e}^{x}})+c\]
E) \[{{e}^{x}}\tan (x{{e}^{x}})+c\]
Correct Answer: C
Solution :
Let \[I=\int{\frac{{{e}^{x}}(1+x)}{{{\cos }^{2}}(x{{e}^{x}})}}dx\] Put \[x{{e}^{x}}=t\] \[\Rightarrow \] \[(x{{e}^{x}}+{{e}^{x}})dx=dt\] \[\Rightarrow \] \[{{e}^{x}}(x+1)dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{{{\cos }^{2}}t}}=\int{{{\sec }^{2}}t\,dt}\] \[=\tan t+c\] \[=\tan (x{{e}^{x}})+c\]
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