# How do you integrate #int (2x^3) / sqrt(1 - x^4) dx#?

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To integrate (\int \frac{{2x^3}}{{\sqrt{{1 - x^4}}}} , dx), we can use the substitution method. Let (u = 1 - x^4), then (du = -4x^3 dx). Rearranging for (dx), we get (dx = -\frac{{du}}{{4x^3}}). Substituting these into the integral, we have:

(\int \frac{{2x^3}}{{\sqrt{{1 - x^4}}}} , dx = \int \frac{{2x^3}}{{\sqrt{{u}}}} \left(-\frac{{du}}{{4x^3}}\right) = -\frac{1}{2}\int \frac{{du}}{{\sqrt{u}}}).

Now, integrating (\int \frac{{du}}{{\sqrt{u}}}) with respect to (u), we get:

(-\frac{1}{2}\int \frac{{du}}{{\sqrt{u}}} = -\frac{1}{2} \cdot 2\sqrt{u} + C = -\sqrt{u} + C).

Finally, substituting back (u = 1 - x^4), we have:

(-\sqrt{1 - x^4} + C).

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