A) \[{{\tan }^{-1}}x+\log (1+{{x}^{2}})+{{\tan }^{-1}}y+\log (1+{{y}^{2}})=c\]
B) \[{{\tan }^{-1}}x-\frac{1}{2}\log (1+{{x}^{2}})+{{\tan }^{-1}}y-\frac{1}{2}\log (1+{{y}^{2}})=c\]
C) \[{{\tan }^{-1}}x+\frac{1}{2}\log (1+{{x}^{2}})+{{\tan }^{-1}}y+\frac{1}{2}\log (1+{{y}^{2}})=c\]
D) None of these
Correct Answer: C
Solution :
Given equation \[(1+{{x}^{2}})(1+y)dy+(1+x)(1+{{y}^{2}})dx=0\] Þ \[\frac{(1+y)}{(1+{{y}^{2}})}dy=-\frac{(1+x)}{(1+{{x}^{2}})}dx\] Þ \[\int_{{}}^{{}}{\left[ \frac{1}{1+{{y}^{2}}}+\frac{y}{1+{{y}^{2}}} \right]}dy+\int_{{}}^{{}}{\left[ \frac{1}{1+{{x}^{2}}}+\frac{x}{1+{{x}^{2}}} \right]dx+c}=0\] Þ \[{{\tan }^{-1}}y+\frac{1}{2}\log (1+{{y}^{2}})+{{\tan }^{-1}}x+\frac{1}{2}\log (1+{{x}^{2}})=c\].
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