Question 15 and question 18 antidifferentiate?

#15: 1/5 (x^2 - x - 3)^5

#18: 5 (3x^2 - 6x + 1)^5

Step-by-step explanation:

#15

(2x-1)(x^2-x+3)^4

Chain Rule!!!

g'(x)f'(g(x)) = d/dx(f(g(x)) = (f ⚬ g)'(x)

f'(x) = x^4

g(x) = x^2 - x - 3

g'(x) = 2x - 1

So we only have to integrate x^4 gives 1/5 x^5

f(x) = 1/5 x^5

f(g(x)) = (f ⚬ g)(x) = 1/5 (x^2 - x - 3)^5

#18

150(x-1)(3x^2-6x+1)^4

Chain rule again!! In disguise.

f'(x) = x^4

f(x) = 1/5 x^5

g(x) = 3x^2-6x+1

g'(x) = 6x - 6 = 6(x-1)

150(x-1) = 25 g'(x)

25 g'(x)f'(g(x)) = 25 (f ⚬ g)(x) = 5(3x^2-6x+1)^5

linear pair postulates