Io farei così ...
$int (2t^2)/(2t-1)\ \ dt=1/2int (4t^2)/(2t-1)\ \ dt=1/2int (4t^2-1+1)/(2t-1)\ \ dt=1/2int ((2t-1)(2t+1)+1)/(2t-1)\ \ dt$
$1/2int ((2t-1)(2t+1))/(2t-1)\ \ dt+1/2int 1/(2t-1)\ \ dt=1/2int (2t+1)\ \ dt+1/2int 1/(2t-1)\ \ dt$
$1/2int 2t \ dt + 1/2int 1\ \ dt+1/4int 1/(t-1/2)\ \ dt$
da cui ...
$t^2/2+t/2+(log|t-1/2|)/4$