I=∫4x+62x2+5x+3dx=∫4x+6dx2x2+3x+2x+3=∫(4x+6)dx(2x+3)x+1(2x+3)
I=∫dx(4x+6)(x+1)(2x+3)=∫dx(4x+4)(x+1)(2x+3)+2∫dx(x+1)(2x+3)
=∫4dx(2x+3)+2∫dx(x+1)(2x+3)
I=a|n|2x+3|=2[2(2x+3)−1(x+1)]
I=4|n|2x+3|−4|n|2x+3|+2|n|x+1|+|n|c|
I=|n|c(x+1)2|
∴∫4x+6(2x2+5x+3)dx=|n|c(x+1)2| c= constant.