Question:

Put $1 + x^2 = t \Rightarrow 2xdx = dt$<br>$\therefore \int \frac{2x^3 dx}{1+x^2} = \int \frac{2x \cdot x^2 dx}{1+x^2} = \int \frac{(t-1)}{t} dt = \int \left(1 - \frac{1}{t}\right) dt$<br>$= t - \ln t + C = (1 + x^2) - \ln(1 + x^2) + C$