A) \[\cos \,(x{{e}^{x}})\]
B) \[sin\,(x{{e}^{x}})\]
C) \[2{{\tan }^{-1}}(x)\]
D) \[\tan \,(x{{e}^{x}})\]
Correct Answer: D
Solution :
| Given that \[\int{{{e}^{x}}\left( 1+x \right).{{\sec }^{2}}\left( x{{e}^{x}} \right)dx=f(x)+\,\,}\text{constant}\] |
| Put \[x{{e}^{x}}=t\] in L.H.S. \[\Rightarrow \,\left( {{e}^{x}}+x{{e}^{x}} \right)dx=dt\] |
| \[\because \] L.H.S. \[=\int{{{\sec }^{2}}\,t\,dt=\tan t+\,}\text{constant}\] |
| \[\Rightarrow \,\,\,\,\tan \,\,(x\,{{e}^{x}})+constant=f(x)+constant\] |
| \[\Rightarrow \,\,f(x)=\tan \,(x{{e}^{x}})\] |
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