Evaluate the integral ∫ x 5 ( x 6 − 3 ) 22 d x by making the substitution u = x 6 − 3

Answer 1

Let u = x⁶ - 3 then du/dx = 6x⁵ or du = 6x⁵ dx

Prior to performing the substitution, you can think of the original problem as:

∫(6/6)x⁵(x⁶ - 3) dx = 1/6∫6x⁵(x⁶ - 3) dx =

1/6∫(x⁶ - 3)6x⁵ dx

Now, perform the substitution (u = x⁶ - 3 and du = 6x⁵ dx) making the integral:

1/6∫u du

Evaluate the integral:

1/6(1/2u² + C) = 1/12u² + C

Now, back substitute (u = x⁶ - 3) to get:

1/12(x⁶ - 3)² + C

Double check by taking the derivative to make sure it matches the original problem by using the power rule and the chain rule and it checks out.