Derivative of a product with e^x combined with a log term

For the first term, apply the product rule to $x e^x$: with $f(x)=x$ and $h(x)=e^x$, we get $\frac{d}{dx}[x e^x] = (1)e^x + x(e^x) = e^x + xe^x$.

For the second term, apply the chain rule to $\ln(x^2+1)$: the outer function is $\ln(u)$ and the inner function is $u=x^2+1$, so $\frac{d}{dx}[\ln(x^2+1)] = \frac{1}{x^2+1}\cdot 2x = \frac{2x}{x^2+1}$. Subtracting this from the first result gives $g'(x) = e^x + xe^x - \frac{2x}{x^2+1}$.