(e^(y)+1)xdx=cxtc1" exdy "

(e^(y)+1)xdx=(x+1)e^(y)dy

(e ^ (y) +1) cos xdx + e ^ (y) sin xdy = 0

The general solution of the equation (e^(y)+1)cos xdx+e^(y)sin xdx=0 is

int(e^(2x)+1)/e^xdx

If e^(x)+e^(y)=e^(x+y)" then "y_(1)=

int(e^x+1)^2e^xdx

e^(int sec xdx)

e^(int tan xdx)

The equation of the curve passing through ((pi)/(6),0) and satisfying the differential equation (e^y+1)cos xdx+e^ysin xdy=0 , is

int_(e)^(e^(2))(log xdx)/((1+log x)^(2))=(e)/(6)(2e-3)