Solve differential equation $(1+x^2) \frac{dy}{dx} - 2xy = x$ Solve differential equation $$(1+x^2) \frac{dy}{dx} - 2xy = x$$
I simplified it to $$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$
$$ \int \frac{1}{1+2y} dy =\int \frac{x}{1+x^2} dx $$
$$\frac{1}{2} \int \frac{2}{1+2y} dy = \frac{1}{2} \int \frac{2x}{1+x^2} dx$$
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $$
From here, I got stuck. I have to remove y from here to solve it.
The answer in the textbook gave $ y= (k(1+x^2) - 1)/{2}$ I believe $k$ is the integration constant. How do I remove the $\ln$ from both sides? 

$$\frac{1}{1+2y} dy = \frac{x}{1+x^2} dx$$ 
$$\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + \frac {1}{2} \ln C$$
$$ | 1 + 2y | =C(1+x^2)$$
$$ 2y = -1 +C(1+x^2)$$
$$ y = \frac {-1 +C(1+x^2)}{2}$$
