∫3x2+2xx6+2x5+x4+2x3+2x2+5dx=
14tan−1[x3+x2+12]+c
12tan−1[x3+x2+12]+c
sin−1[x3+x2+12]+c
12tan−1[x3+x22]+c
∫3x2+2x(x3+x2+1)2+22dx
Let x3+x2+1=t
∫dtt2+22=12tan−1(t2)+c
=12tan−1[x3+x2+12]+c