What is the solution for y^'=(-1+x+y)/(1+x) ?

The solution for y^'=(-1+x+y)/(1+x) is y=2+ln(x+1)(1+x)+c_{1}(1+x)

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y^'=(-1+x+y)/(1+x)

y^{^{'}} = \frac{- 1 + x + y}{1 + x}

y = 2 + ln (x + 1) (1 + x) + c_{1} (1 + x)

Solution steps

y^{'} (x) = \frac{- 1 + x + y}{1 + x}

Solve linear ODE:

y = 2 + ln (x + 1) (1 + x) + c_{1} (1 + x)

y = 2 + ln (x + 1) (1 + x) + c_{1} (1 + x)

\frac{9 - ln ( x )}{ln ( x ) - 1}

\int_{0}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

2 x^{2}, 8 x - 2 x^{2}

(2 x - 1) d x + (5 y + 3) d y = 0

\int 1 + \frac{16}{x ^{\frac{2}{3}}} d x

What is the solution for y^'=(-1+x+y)/(1+x) ?
The solution for y^'=(-1+x+y)/(1+x) is y=2+ln(x+1)(1+x)+c_{1}(1+x)