Consider the expression, f(2t+1t+1)=t+1 Take, 2t+1t+2=xt=2x−11−x This implies, f(x)=2x−11−x+1f(x)=2x−11−x+2x−1f(x)=2x−1x So, ∫f(x)dx=∫2x−1xdx=21∫2x−12xdx=21∫2x−12x−1+1dx=21[x+21log(2x−1)]+c Further simplify the above, ∫f(x)dx=2x+41log(2x−1)+c