A) \[\frac{1}{2}x{{e}^{{{\tan }^{-1}}x}}+C\]
B) \[x.{{e}^{{{\tan }^{-1}}x}}+C\]
C) \[\frac{1}{2}.{{e}^{{{\tan }^{-1}}x}}+C\]
D) \[{{e}^{{{\tan }^{-1}}x}}+C\]
Correct Answer: B
Solution :
[b] \[I=\int{{{e}^{{{\tan }^{-1}}x}}}\left( 1+\frac{x}{1+{{x}^{2}}} \right).dx=\int{{{e}^{{{\tan }^{-1}}x}}}.dx+\int{{{e}^{{{\tan }^{-1}}x}}}.\frac{x}{1+{{x}^{2}}}.dx\]\[={{e}^{{{\tan }^{-1}}x}}.x-\int{x.{{e}^{{{\tan }^{-1}}x}}}.\frac{1}{1+{{x}^{2}}}.dx+\int{{{e}^{{{\tan }^{-1}}x}}}.\frac{x}{1+{{x}^{2}}}.dx+c\]\[=x.{{e}^{{{\tan }^{-1}}x}}+c\] Hence, option [b] is correct.
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