A) \[\frac{1}{2}\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]
B) \[\frac{1}{2}\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]
C) \[\log (1+{{y}^{2}})=\log x-{{\tan }^{-1}}x+c\]
D) \[\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\]
Correct Answer: B
Solution :
\[xy\frac{dy}{dx}=\frac{(1+{{y}^{2}})(1+x+{{x}^{2}})}{(1+{{x}^{2}})}\] Þ \[\int_{{}}^{{}}{\frac{ydy}{1+{{y}^{2}}}=\int_{{}}^{{}}{\frac{(1+x+{{x}^{2}})}{x(1+{{x}^{2}})}}}dx=\int_{{}}^{{}}{\frac{1}{x}dx}+\int_{{}}^{{}}{\frac{dx}{1+{{x}^{2}}}}\] Þ \[\frac{1}{2}\log (1+{{y}^{2}})=\log x+{{\tan }^{-1}}x+c\].
You need to login to perform this action.
You will be redirected in
3 sec
