Use integration by parts to find \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
Use the substitution \(t = 2 x + 1\) to show that \(\int 4 x \ln ( 2 x + 1 ) \mathrm { d } x\) can be written as \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
Hence find the exact value of \(\int _ { 0 } ^ { 1 } 4 x \ln ( 2 x + 1 ) \mathrm { d } x\).