∫ 2x12+5x9dx1+x3+x53=
x102x5+x3+12+c
x2+2xx5+x3+1+c
lnx5+x3+1+2x12+5x9+c
lnx5+x3+1+2x12−5x9+c
I=∫ 2x12+5x91+x3+x53dx=∫ 5x−6+2x−3dx1+x−2+x−53
Putting 1+x-2+x-5=t , then
−2x−3−5x−6dx=dt
∴ I=−∫ dtt3=12t2=121+x−2+x−52
=x102x5+x3+12+c