A) \[\frac{1}{2}\log (x-1)-\frac{1}{4}\log ({{x}^{2}}+1)-\frac{1}{2}{{\tan }^{-1}}x+c\]
B) \[\frac{1}{2}\log (x-1)+\frac{1}{4}\log ({{x}^{2}}+1)-\frac{1}{2}{{\tan }^{-1}}x+c\]
C) \[\frac{1}{2}\log (x-1)-\frac{1}{2}\log ({{x}^{2}}+1)-\frac{1}{2}{{\tan }^{-1}}x+c\]
D) None of these
Correct Answer: A
Solution :
We have \[\frac{1}{(x-1)({{x}^{2}}+1)}=\frac{A}{(x-1)}+\frac{Bx+C}{({{x}^{2}}+1)}\] \[\Rightarrow 1=A({{x}^{2}}+1)+(Bx+C)(x-1)\] If \[x=1,\] then \[A=\frac{1}{2}\] .....(i) \[A-C=1\Rightarrow C=-\frac{1}{2}\] .....(ii) \[A+B=0\Rightarrow B=-\frac{1}{2}\] .....(iii) Putting these values, we get \[\frac{1}{(x-1)({{x}^{2}}+1)}=\frac{1}{2}.\frac{1}{(x-1)}-\frac{x+1}{2({{x}^{2}}+1)}\] Hence \[\int_{{}}^{{}}{\frac{1}{(x-1)({{x}^{2}}+1)}}dx=\frac{1}{2}\int_{{}}^{{}}{\frac{dx}{(x-1)}-\frac{1}{2}}\,\int_{{}}^{{}}{\frac{x+1}{{{x}^{2}}+1}}dx\] \[=\frac{1}{2}\log (x-1)-\frac{1}{4}\log ({{x}^{2}}+1)-\frac{1}{2}{{\tan }^{-1}}x+c.\]
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