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/x³first, see that 32/x³= 32 * x^-3. Which we integrate as such -32/2 * x^-2because 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^-2were 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.)

7^6 x 7^3

Related Maths A Level answers

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

A curve is described by f(x) = x^2 + 2x. A second curve is described by g(x) = x^2 -5x + 7. Find the point (s) where both curves intersect.

Solve for 0<=θ<π, the equation sin3θ-(sqrt3)cosθ=0 (C2)

How do I integrate by substitution?