Two different ways of substitution leads to two different results for $\int\frac{x\ln x^2+1}{x^2+1}dx$, how's that? I was working on a solution for
$$\int\frac{x\ln (x^2+1)}{x^2+1}dx$$ and my approach was the following:
first substituting $u = x^2+1$, which means $du = 2x dx$, and therefore:
$$\int\frac{\ln u}{2u}du$$
and then substituting $v=\ln u$, which means $dv = \frac{1}{u} du$, and therefore:
$$\frac{1}{2}\int dv = \frac{v}{2} + c = \frac{\ln u}{2} +c = \frac{\ln (x^2+1)}{2}+c$$
However verifying this online from various sources, they indicate that the solution is in fact:
$$\int\frac{x\ln (x^2+1)}{x^2+1}dx = \frac{\ln(x^2+1)^2}{4}+c$$
because they substitute $u = \ln(x^2+1)$ which gives $du = \frac{2x}{x^2+1} dx$, which is clairly a shorter path.
But my question is, why does both of these substitution techniques not yield the same result? Where lies my error?

$$I=\int\frac{x\ln(x^2+1)}{x^2+1}dx$$
$$u=x^2+1$$
$$dx=\frac{du}{2x}$$
$$I=\frac{1}{2}\int\frac{\ln(u)}{u}du$$
$$v=\ln(u)$$
$$du=udv$$
$$I=\frac{1}{2}\int vdv=\frac{v^2}{4}+C=\frac{\ln^2(u)}{4}+C=\frac{\ln^2(x^2+1)}{4}+C$$
