A) \[y=\tan ({{x}^{2}}+x+c)\]
B) \[y=\tan (2{{x}^{2}}+x+c)\]
C) \[y=\tan ({{x}^{2}}-x+c)\]
D) \[y=\tan \left( \frac{{{x}^{2}}}{2}+x+c \right)\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}=(1+x)(1+{{y}^{2}})\] Þ \[\frac{dy}{1+{{y}^{2}}}=(1+x)dx\] On integrating both sides, we get \[{{\tan }^{-1}}y=\frac{{{x}^{2}}}{2}+x+c\] Þ \[y=\tan \left( \frac{{{x}^{2}}}{2}+x+c \right)\].You need to login to perform this action.
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