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Answer 1 · 2017-04-05T00:57:57

Use the domain of the integrand and use #sqrt(x^2) = abs(x)# and integrate by substitution.

I am assuming that #a > 0#

Depending on the domain of integration, we can integrate

#int sqrt(x^2(x^2-a)) dx = int absx sqrt(x^2-a) dx# by substitution.

Note that the domain of the integrand is #(-oo,-sqrta] uu [sqrta,oo)#

On #(-oo,-sqrta]#, we have #sqrt(x^2(x^2-a)) = -xsqrt(x^2-a)#

So

#int sqrt(x^2(x^2-a)) = int -xsqrt(x^2-a) = -1/3(x^2-a)^(3/2) +C#

On #[sqrta,oo)#, we have #sqrt(x^2(x^2-a)) = xsqrt(x^2-a)#

So

#int sqrt(x^2(x^2-a)) = int xsqrt(x^2-a) = 1/3(x^2-a)^(3/2) +C#

Note

Because #sqrt(x^2)/x = absx/x = {(1,"if",x < 0),(-1,"if",x < 0):}#, we could write the answer as

#sqrt(x^2)/(3x)(x^2-a)^(3/2) +C# or as

#(x^2(x^2-a))^(3/2)/(3x^3)#