need help with exact differential equation?

Solve this differential equation by using an integrating factor:

(1 - x²)(dy / dx) - x²y = (1 + x)√(1 - x²)

dy / dx - x²y / (1 - x²) = (1 + x) / √(1 - x²)

dy / dx + x²y / (x² - 1) = (1 + x) / √(1 - x²)

P(x) = x² / (x² - 1)

f(x) = (1 + x) / √(1 - x²)

I(x) = ℮^[∫ P(x) dx]

I(x) = ℮^[∫ x² / (x² - 1) dx]

I(x) = ℮^[∫ {1 - 1 / (1 - x²)} dx]

I(x) = ℮^(x - tanhˉ¹x)

I(x)y = ∫ I(x)f(x) dx

y℮^(x - tanhˉ¹x) = ∫ (1 + x)℮^[x - tanhˉ¹x] / √(1 - x²) dx

y℮^(x - tanhˉ¹x) = ∫ ℮˟ dx

y℮^(x - tanhˉ¹x) = ℮˟ + C

y℮^(x - tanhˉ¹x) = C + ℮˟

y = (C + ℮˟) / ℮^(x - tanhˉ¹x)

(1−x²)dy/dx − x²y = (1+x)√(1−x²)

Divide by 1−x² : dy/dx − {x²/(1−x²)}y = √{(1+x)/(1−x)} … (i)

IF is exp{ ∫ −{x²/(1−x²)}dx }

∫ {x²/(1−x²)}dx = ∫ { 1/(1−x²) − 1 }dx = ∫ { ½/(1+x) + ½/(1−x) − 1 } dx

= ½log(1+x) − ½log(1−x) − x = ½log{(1+x)/(1−x)} − x

â��´ IF = exp{ x−½log{(1+x)/(1−x)} } = exp(x)*√{(1−x)/(1+x)}

Multiplying (i) by IF, noting that the LHS is always d( IF*y )/dx

d( y*exp(x)*√{(1−x)/(1+x)} )/dx = exp(x)

Integrating : y*exp(x)*√{(1−x)/(1+x)} = exp(x) + c

∴ y = √{(1+x)/(1−x)} * (1+cexp(−x) )