Substitute:
#x = 2tant#
#dx = 2sec^2tdt#
with #t in (-pi/2,pi/2)#, so that:
#int x^3sqrt(x^2+4)dx = 2 int (2tant)^3 sqrt((2tant)^2+4) sec^2t dt#
#int x^3sqrt(x^2+4)dx = 32 int tan^3t sqrt(tan^2t+1) sec^2t dt#
Use now the trigonometric identity:
#tan^2t+1 = sec^2t#
and as for #t in (-pi/2,pi/2)# the secant is positive:
#sqrt(tan^2t+1) = sect#
so:
#int x^3sqrt(x^2+4)dx = 32 int tan^3t sec^3 dt#
Now:
#tan^3t = tant * tan^2t = tant(sec^2t-1)#,
then:
#int x^3sqrt(x^2+4)dx = 32 int tant (sec^2t-1)sec^3 dt#
and using the linearity of the integral:
#int x^3sqrt(x^2+4)dx = 32 int tant sec^5t dt -32 int tant sec^3tdt#
and as:
#d/dt sect = sect tant#
we have:
#int x^3sqrt(x^2+4)dx = 32 int sec^4t d(sect) -32 int sec^2t d(sect)#
#int x^3sqrt(x^2+4)dx = 32/5 sec^5t -32/3 sec^3t +C#
Now to undo the substitution:
#sect = sqrt(tan^2+1) = sqrt((x/2)^2+1) = sqrt(x^2+4)/2#
so:
#int x^3sqrt(x^2+4)dx = (x^2+4)^(5/2)/5 -4/3 (x^2+4)^(3/2) +C#
