A) \[\log [x\,{{(1-y)}^{2}}]=\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{2}-2y+c\]
B) \[\log [x{{(1-y)}^{2}}]=\frac{{{x}^{2}}}{2}-\frac{{{y}^{2}}}{2}+2y+c\]
C) \[\log [x{{(1+y)}^{2}}]=\frac{{{x}^{2}}}{2}+\frac{{{y}^{2}}}{2}+2y+c\]
D) \[\log [x{{(1-y)}^{2}}]=\frac{{{x}^{2}}}{2}-\frac{{{y}^{2}}}{2}-2y+c\]
Correct Answer: D
Solution :
\[(1-{{x}^{2}})(1-y)dx=xy(1+y)dy\] Þ \[\int_{{}}^{{}}{\frac{y(1+y)}{(1-y)}dy}=\int_{{}}^{{}}{\frac{(1-{{x}^{2}})}{x}dx}\]; Now integrate it.You need to login to perform this action.
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