# How do you integrate #int-16x^3(-4x^4-1)^-5# using substitution?

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To integrate ( \int 16x^3(-4x^4 - 1)^{-5} ) using substitution, let ( u = -4x^4 - 1 ). Then, ( du = -16x^3 dx ). Substituting these into the integral, we get ( -\frac{1}{16} \int u^{-5} du ). Integrating ( u^{-5} ), we have ( -\frac{1}{16} \cdot \frac{u^{-4}}{-4} + C ). Substituting back ( u = -4x^4 - 1 ), the final result is ( \frac{1}{64x^4 + 16} + C ).

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