What is the solution for (1+x)^2y^'+(1+x)y=1 ?

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

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(1+x)^2y^'+(1+x)y=1

(1 + x)^{2} y^{^{'}} + (1 + x) y = 1

y = \frac{ln ( 1 + x )}{1 + x} + \frac{c _{1}}{1 + x}

Solution steps

(1 + x)^{2} y^{'} (x) + (1 + x) y = 1

Solve linear ODE:

y = \frac{ln ( 1 + x )}{1 + x} + \frac{c _{1}}{1 + x}

y = \frac{ln ( 1 + x )}{1 + x} + \frac{c _{1}}{1 + x}

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