A) \[-x\]
B) \[x\]
C) \[\sqrt{1-x}\]
D) \[\sqrt{1+x}\]
Correct Answer: B
Solution :
Let \[I=\int{\frac{{{x}^{2}}-x+1}{{{x}^{2}}+1}}.{{e}^{{{\cot }^{-1}}x}}dx\] Put \[x=\cot \,t\Rightarrow -\cos e{{c}^{2}}t\,dt=dx\] Now, \[1+{{\cot }^{2}}t=\cos e{{c}^{2}}t\] \[\therefore \] \[I=\int{\frac{{{e}^{t}}({{\cot }^{2}}t-\cot \,t+1)}{(1+{{\cot }^{2}}t)}(-\cos e{{c}^{2}}t)}dt\] \[=-\int{{{e}^{t}}(\cos e{{c}^{2}}t-\cot \,t)dt}\] \[=\int{{{e}^{t}}(cot\,t-\cos e{{c}^{2}}t)}\,dt\] \[={{e}^{t}}\cot \,t+C\] \[={{e}^{{{\cot }^{-1}}x}}(x)+C\equiv A(x).{{e}^{{{\cot }^{-1}}x}}+C\] \[\Rightarrow \]\[A(x)=x\]You need to login to perform this action.
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