A) \[\ln \sqrt{1+{{y}^{2}}}=\ln (x)+{{\tan }^{-1}}(x)-\frac{\pi }{2}\]
B) \[\ln \left( \frac{1+{{y}^{2}}}{{{x}^{2}}} \right)=2{{\tan }^{-1}}(x)-\frac{\pi }{2}\]
C) \[\ln \left( \frac{1+{{y}^{2}}}{{{x}^{2}}} \right)=\frac{\pi }{4}-2{{\tan }^{-1}}x\]
D) \[\ln \left( \frac{1+{{y}^{2}}}{{{x}^{2}}} \right)={{\tan }^{-1}}(x)-\frac{\pi }{4}\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{ydy}{1+{{y}^{2}}}\,=\int_{{}}^{{}}{\frac{1+x+{{x}^{2}}}{x(1+{{x}^{2}})}dx;}}\] \[\Rightarrow \,\,\frac{1}{2}\,\ln \,(1+{{y}^{2}})\,={{\tan }^{-1}}\,x+\ln x\,+C\] \[\Rightarrow \,\ln \,\frac{1+{{y}^{2}}}{{{x}^{2}}}=2\,{{\tan }^{-1}}\,x+C\]
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