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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

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