Question 565519
By a change of variables, determine the indefinite integral as follows:

integral (x^2 - 3x^4)^1/2 dx
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The integrand can be factored as:
{{{sqrt(x^2(1 - 3x^2)) = x*sqrt(1-3x^2)}}}
Let {{{y = 1 - 3x^2}}}
Then {{{dy = -6x*dx}}}
So the integral can be written as
{{{(-1/6)*int(y^(1/2),dy)}}}
The antiderivative of {{{(-1/6)y^(1/2) = (-1/6)(2/3)*y^(3/2)}}} 
Substituting back the expression for y gives
{{{-(1-3x^2)^(3/2)/9}}}