Let u²=1+x², then 2udu=2xdx, udu=xdx.
x⁴=u⁴-2u²+1
∫√(1+x²)x⁵dx=∫√(1+x²)x⁴xdx=∫u(u⁴-2u²+1)udu.
∫(u⁶-2u⁴+u²)du=u⁷/7-2u⁵/5+u³/3+c=
√(1+x²)((1+x²)³/7-2(1+x²)²/5+(1+x²)/3)+c, where c is integration constant.
∫secxdx=∫(secx(secx+tanx)/(secx+tanx))dx=
∫(sec²x+secxtanx)/(secx+tanx).dx=
ln|A(secx+tanx)| where A is constant.
[If u=(secx+tanx), then du=(secxtanx+sec²x)dx=usecxdx, secxdx=du/u which when integrated is ln|u|.]