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