$$\int \ln(x^2 +1)\ dx$$

I done it using integration by parts where 

$\int u\ dv = uv - \int v\ du$

Let $u$ = $\ln(x^2 +1)$ 

$du = \frac{2x}{x^2+1} dx $

Let $dv = dx$ so $v=x$ 

$\int \ln(x^2 +1)\ dx = x \ln (x^2 +1) - \int \frac{2x^2}{x^2+1} $

I integrate $2\int \frac{x^2}{x^2+1} $ separately using substitution.

Let $u$ = $x^2 + 1$ -> $dx= \frac{1}{2x} du $

substitute $u$ back into it I get a simplified $\int \frac{x}{x^2 + 1} dx$ 

I used substitution again and got $ \frac{1}{2} \int \frac{1}{u} du$

When I sub it back into the original question, this is my final answer, 

$x\ln(x^2 +1) + \frac{1}{2} \ln(x^2 +1) + C$

However, this answer is wrong, 

The answer is, $x\ln(x^2 +1) - 2x + 2\tan^{-1} (x) + C $

I believe my integration of $2\int \frac{x^2}{x^2+1} $ is wrong. Where did I went wrong ?