Find the indefinite integral of ( 32/(x^3) + bx) over x for some constant b.

To integrate (32/ x³ + bx) over x we can integrate each part separately. This is because addition of two integrals is the same as integrating the total of the elements. 

So we integrate 32/x³ first, see that 32/x³ = 32 * x^-3. Which we integrate as such -32/2 * x^-2  because when we integrate x the power is raised by 1 so -3 goes to -2. Now we must find the new coefficient such that when x^-2 were differentiated we get 32 as the resulting coefficient for x^-3. This gives the new coefficient -32/2 as -32/2*-2=32. This gives the integral as -16/x². Now integrate bx. Given the power is raised by 1 the new power is 2 and then to find the new coefficient we must find coeff such that coeff*2 = b. This means that the new coefficient is b/2. This gives the integral as b/2 *x².

Now we add them back together to find the original integral we get -16/x² + b/2 *x² +c. Don't forget the +c! (We need the constant becuase it is an indefinite integral meaning there are no bounds.)