Substitute #x = y^2#, #dx = 2ydy#:
#int y(y^4+y^2+4)^(1/2) dy = 1/2 int sqrt(x^2+x+4)dx#
Complete now the square under the root:
#int y(y^4+y^2+4)^(1/2) dy = 1/2 int sqrt((x+1/2)^2+15/4)dx#
#int y(y^4+y^2+4)^(1/2) dy = sqrt15/4 int sqrt(((2x+1)/sqrt15)^2+1)dx#
and substitute:
#(2x+1)/sqrt15 = tant#
#dx = sqrt15/2 sec^2t dt#
with #t in (-pi/2,pi/2)#
so that:
#int y(y^4+y^2+4)^(1/2) dy =15/8 int sqrt(tan^2t+1)sec^2t dt#
Using the trigonometric identity:
#tan^2t+1 = sec^2t#
and considering that for # t in (-pi/2,pi/2)# the secant is positive:
#int y(y^4+y^2+4)^(1/2) dy =15/8 int sec^3t dt#
Solve now the resulting integral by parts:
#int sec^3t dt = int sect d/dt (tant) dt#
#int sec^3t dt = sect tant - int tant d/dt (sect) dt#
#int sec^3t dt = sect tant - int sect tan^2t dt#
ans using the same identity as above:
#int sec^3t dt = sect tant - int sect (sec^2t -1) dt#
#int sec^3t dt = sect tant - int sec^3tdt +int sect dt#
the same integral now appears on both sides of the equation:
#2int sec^3t dt = sect tant +ln abs (sect +tant) +C#
#int sec^3t dt = (sect tant)/2 +1/2 ln abs (sect +tant) +C#
so:
#int y(y^4+y^2+4)^(1/2) dy =15/16 (sect tant) +15/16 ln abs (sect +tant) +C#
To undo the substitution note that:
#tant = (2x+1)/sqrt15#
#sect = sqrt(tan^2t +1) = sqrt( (2x+1)^2/15 +1) = sqrt(( 4x^2+4x+16)/15) = 2/sqrt15 sqrt(x^2+x+4)#
so:
#int y(y^4+y^2+4)^(1/2) dy =15/16 ((2x+1)/sqrt15) (2/sqrt15 sqrt(x^2+x+4)) + 15/16ln abs ( (2x+1)/sqrt15 + 2/sqrt15 sqrt(x^2+x+4)) +C#
#int y(y^4+y^2+4)^(1/2) dy = ((2x+1) sqrt(x^2+x+4))/8 +15/16 ln abs ( (2x+1)+ 2 sqrt(x^2+x+4)) +C#
and finally:
#int y(y^4+y^2+4)^(1/2) dy = ((2y^2+1) sqrt(y^4+y^2+4))/8 + 15/16ln abs ( (2y^2+1)+ 2 sqrt(y^4+y^2+4)) +C#
