Question

determine the integral of function ( x^3 -2 )(x^4 -8x)^0.5 dx

Answer 1

Let us determine the integral

$\int( x^3 -2 )(x^4 -8x)^{0.5} dx\\ =\frac{1}{4}\int( 4x^3 -8 )(x^4 -8x)^{0.5} dx\\ =\frac{1}{4}\int(x^4 -8x)^{0.5} d( x^4 -8x)\\ =\frac{1}{4}\frac{2}{3}(x^4 -8x)^{\frac{3}{2}}+C\\ =\frac{1}{6}(x^4 -8x)^{\frac{3}{2}}+C$