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? 

You got : $\quad\frac{1}{2} \ln | 1 + 2y | = \frac{1}{2} \ln | 1+x^2 | + C $ 
$$\frac{1}{2} \ln | 1 + 2y | - \frac{1}{2} \ln | 1+x^2 | = C  $$
$$\ln | 1 + 2y | - \ln | 1+x^2 | = 2C  $$
$$\ln \frac{| 1 + 2y |}{ | 1+x^2 |} = 2C  $$
Thus $\quad\frac{ 1 + 2y }{  1+x^2 }=$constant. 
$$\frac{ 1 + 2y }{  1+x^2 }=c$$
$$1+2y=c(1+x^2)$$
$$y=\frac12\left(c(1+x^2)-1\right)$$
