What is the solution for (dy)/(dx)y=(y+1)(2x+1),y(-1)=1 ?

The solution for (dy)/(dx)y=(y+1)(2x+1),y(-1)=1 is x^2+x=y-ln(y+1)+ln(2)-1

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(dy)/(dx)y=(y+1)(2x+1),y(-1)=1

\frac{d y}{d x} y = (y + 1) (2 x + 1), y (- 1) = 1

x^{2} + x = y - ln (y + 1) + ln (2) - 1

Solution steps

\frac{d y}{d x} y = (y + 1) (2 x + 1)

Solve separable ODE:

x^{2} + x = y - ln (y + 1) + ln (2) - 1

x^{2} + x = y - ln (y + 1) + ln (2) - 1

\int z^{2} (z^{3} - 1)^{6} d z

\int p (p + 6)^{5} d p

\int \frac{2}{1 - 4 x ^{2}} d x

f (x) = 2 x - x^{2}, at x = 1

\frac{d}{d x} (\frac{x ^{3} - 1}{ln ( x )})

What is the solution for (dy)/(dx)y=(y+1)(2x+1),y(-1)=1 ?
The solution for (dy)/(dx)y=(y+1)(2x+1),y(-1)=1 is x^2+x=y-ln(y+1)+ln(2)-1