how do i intergrate (3x^3+1)/(x^3+x+7)^1/2?

Answer 1

3/2*x
+1/2*sum((-20-3*r)/(3*r^2+1)
*ln(x-r);r^3+r+7=0))
+C

Answer 2

I think you are asking for Int (3x^2+1)/x^3+3x+7)^1/2 dx
if this is right
z= x^3+x+7so dz = (3x^2+1)dx and the integral becomes
Int z^-1/2 dz = 2z^1/2+c = 2(x^3+x+7)^1/2+C