Question 1117235
Let {{{x=2sec(u)}}}
{{{x^2=4sec^2(u)}}}
{{{dx=2sec(u)tan(u)du}}}
So,
{{{sqrt((x^2-4)^3)dx=
(sqrt(4sec^2(u)-4))^3*2sec(u)tan(u)du}}}
{{{sqrt((x^2-4)^3)dx=
2(sqrt(4tan^2(u)))^3*sec(u)tan(u)du}}}
{{{sqrt((x^2-4)^3)dx=
2(2^3*tan^3(u))*sec(u)tan(u)du}}}
{{{sqrt((x^2-4)^3)dx=
16tan^4(u)*sec(u)du}}}
{{{sqrt((x^2-4)^3)dx=
16(sec^2(u)-1)^4*sec(u)du}}}
{{{sqrt((x^2-4)^3)dx=
16(sec^8(u)-4sec^6(u)+6sec^4(u)-4sec^2(u))(sec(u))du}}}
{{{sqrt((x^2-4)^3)dx=
16(sec^9(u)-4sec^7(u)+6sec^5(u)-4sec^3(u))du}}}
You can now integrate by parts or use partial fractions.