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.)
