Easy calc problem!?

Question

Find the indefinite integral of (x-1) (2-x)^(1/2)dx

Answer 1

The function is (x - 1)(2 - x)^1/2 .

Indefinite integral of the function :

 ∫ [(x - 1)(2 - x)^1/2 ] dx

Let, 2 - x = u ⇒ x = 2 - u

- dx = du

= - ∫ [(2 - u - 1)(u)^1/2 ] du

= - ∫ [(1 - u)(u)^1/2 ] du

= - ∫ (u^1/2 - u^3/2) du

= - ∫ u^1/2 du + ∫ u^3/2 du

Use the power formula : ∫ xⁿ dx = [(x^{n + 1})/(n + 1)] + C.

= ( -  u^{1/2 + 1})/(1/2 + 1) + (u^{3/2 + 1})/(3/2 + 1) + C

= - 2u^3/2/3 + 2u^5/2/5 + C.

Put, u = 2 - x.

= - 2(2 - x)^3/2/3 + 2(2 - x)^5/2/5 + C

= [- 10(2 - x)^3/2 + 6(2 - x)^5/2]/15 + C

= [- 10(2 - x)^3/2 + 6(2 - x)^3/2(2 - x)]/15 + C

= (-2/15)(2 - x)^3/2[5 - 3(2 - x)] + C

= (-2/15)(2 - x)^3/2[5 - 6 + 3x] + C

= (- 2/15)(2 - x)^3/2(3x - 1) + C.

∴  Indefinite integral of (x - 1)(2 - x)^1/2 is (- 2/15)(2 - x)^3/2(3x - 1) + C.