Find the general solution of the differential equations:`(1+x^2)dy+2x y dx=cotx dx(x!=0)`

Explanation

`(1+x^(2))dy+2xydx=cotxdx`
`implies (dy)/(dx)+(2x)/(1+x^(2))y=(cotx)/(1+x^(2))`
Here, `P=(2x)/(1+x^(2))` , `Q=(cotx)/(1+x^(2))`
`:. I.F.=e^(int(2x)/(1+x^(2)))dx=e^(log(1+x^(2)))=(1+x^(2))`
and general solution
`y(1+x^(2))=int(cotx)/(1+x^(2))(1+x^(2))dx+c`
`=intcotxdx+c`
`implies y(1+x^(2))=logsinx+c`