Answer 1

 (1 +x² ) dy + 2xy dx = cot x dx  

(1 +x² ) `dy/dx ` + 2xy = cot x

`dy/dx +(2xy)/(1+x^2) = (cot x)/(1 +x^2)`

`dy/dx + py = q`

Comparing with linear differential equation

I .F . = `e^(intp  dx)`

I.F = `e^(int x/(1+x^2) dx)`

I .F = `int (2x)/(1+x^2) dx`

put  t = 1 +x² 

`dt/dx = 2x`

dt = 2x dx 

`I = int dt/t = "In " t = "In" ( 1+ x^2 )`

`⇒ I .f = e^("in"^((1+x^2) ) = 1 +x^2`

`x(I.F) = int  Q ( I .F) dx + c`

`x(1 +x^2 ) = int (cot x ) /(1 +x^2 )  (1+ x^2 )  dx + c `

` x(1 + x^2 ) = int  cot x  dx + c`

x + x³ = In | sin x | + c

x + x³ - In | sin x | + c + 0