If $y=x\ln (1+x)-2x+2\arctan x$ how I can get $y'=\ln (1+x^2)$?

Question

I know that $\displaystyle\int ln(1+x^2)\,dx=x\ln (1+x)-2x+2\arctan x+c$ but if I derive the primitive function I get $\displaystyle F'(x)=\ln (x+1)+\frac{x}{1+x}-2+\frac{2}{1+x^2}$, It is true that $\displaystyle\ln (x+1)+\frac{x}{1+x}-2+\frac{2}{1+x^2}=\ln(1+x^2)$? Thanks very much.

Answer 1

It is

$$\displaystyle\int ln(1+x^2)\,dx=x\ln (1+\color{red}{x^2})-2x+2\arctan x+c.$$

And if $$F(x)=x\ln (1+x^2)-2x+2\arctan x+c$$ we get

$$F'(x)=\ln(1+x^2)+\frac{2x^2}{1+x^2}-2+\frac{2}{1+x^2}.$$ But

$$\frac{2x^2}{1+x^2}-2+\frac{2}{1+x^2}=0.$$