ddxln(x2+1)=\dfrac{d}{dx}\ln(x^2 + 1) =dxdln(x2+1)=Aln(2x)\ln(2x)ln(2x)B2xln(x2+1)\dfrac{2x}{\ln(x^2+1)}ln(x2+1)2xC2xx2+1\dfrac{2x}{x^2 + 1}x2+12xcheck_circleD1x2+1\dfrac{1}{x^2 + 1}x2+11Explanation1x2+1⋅2x=2xx2+1\dfrac{1}{x^2+1} \cdot 2x = \dfrac{2x}{x^2+1}x2+11⋅2x=x2+12x.