A) \[x{{e}^{{{\tan }^{-1}}}}x+c\]
B) \[{{\tan }^{-1}}x+C\]
C) \[{{e}^{{{\tan }^{-1}}x}}+2x+C\]
D) None of these
Correct Answer: A
Solution :
[a] Put \[x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta d\theta \] \[I=\int{{{e}^{\theta }}\frac{1+\tan \theta +{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }.{{\sec }^{2}}\theta d\theta }\] \[=\int{{{e}^{\theta }}(tan\theta +se{{c}^{2}}\theta )d\theta }\] \[={{e}^{\theta }}\tan \theta +c=x{{e}^{{{\tan }^{-1}}x}}+c\]
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