Integral by parts?

Hello,

let's write the integral (being d(x - 1) the same as dx) as:

∫ x² (x - 1)⁵ d(x - 1) =

let's integrate by parts, letting:

x² = u → 2x dx = du

(x - 1)⁵ d(x - 1) = dv → [1/(5+1)] (x - 1)⁵ ⁺ ¹ = (1/6)(x - 1)⁶ = v

yielding:

∫ u dv = u v - ∫ v du

∫ x² (x - 1)⁵ d(x - 1) = x² (1/6)(x - 1)⁶ - ∫ (1/6)(x - 1)⁶ 2x dx =

(simplifying)

(1/6)x² (x - 1)⁶ - ∫ (1/3)x (x - 1)⁶ d(x - 1) =

let's integrate by parts further, letting:

(1/3)x = u → (1/3) dx = du

(x - 1)⁶ d(x - 1) = dv → [1/(6+1)] (x - 1)⁶ ⁺ ¹ = (1/7)(x - 1)⁷ = v

obtaining:

∫ u dv = u v - ∫ v du

(1/6)x² (x - 1)⁶ - [(1/3)x (1/7)(x - 1)⁷ - ∫ (1/7)(x - 1)⁷ (1/3) dx] =

(1/6)x² (x - 1)⁶ - (1/21)x (x - 1)⁷ + ∫ (1/21)(x - 1)⁷ dx =

(factoring the constant out)

(1/6)x² (x - 1)⁶ - (1/21)x (x - 1)⁷ + (1/21) ∫ (x - 1)⁷ d(x - 1) =

(1/6)x² (x - 1)⁶ - (1/21)x (x - 1)⁷ + (1/21) [1/(7+1)] (x - 1)⁷ ⁺ ¹ + C =

(1/6)x² (x - 1)⁶ - (1/21)x (x - 1)⁷ + (1/21)(1/8)(x - 1)⁸ + C =

(1/6)x² (x - 1)⁶ - (1/21)x (x - 1)⁷ + (1/168)(x - 1)⁸ + C

if we want to turn this into your answer, let's do as follows:

let's factor out (x - 1)⁶:

[(1/6)x² - (1/21)x (x - 1) + (1/168)(x - 1)²] (x - 1)⁶ + C =

{[28x² - 8x (x - 1) + (x - 1)²] /168} (x - 1)⁶ + C =

[(28x² - 8x² + 8x + x² - 2x + 1) /168] (x - 1)⁶ + C =

[(21x² + 6x + 1) /168] (x - 1)⁶ + C =

let's subract and add 42x (that is 2(21x) ) in the numerator:

[(21x² - 42x + 42x + 6x + 1) /168] (x - 1)⁶ + C =

[(21x² - 42x + 48x + 1) /168] (x - 1)⁶ + C =

let's add and subtract 21 in the numerator:

[(21x² - 42x + 21 + 48x + 1 - 21) /168] (x - 1)⁶ + C =

(grouping terms)

{[(21x² - 42x + 21) + 48x - 20] /168} (x - 1)⁶ + C =

(factoring 21 out of the trinomial)

{[21(x² - 2x + 1) + 48x - 20] /168} (x - 1)⁶ + C =

let's subtract and add 48 in the numerator:

{[21(x² - 2x + 1) + 48x - 48 + 48 - 20] /168} (x - 1)⁶ + C =

{[21(x - 1)² + (48x - 48) + 28] /168} (x - 1)⁶ + C =

(factoring 48 out of the binomial)

{[21(x - 1)² + 48(x - 1) + 28] /168} (x - 1)⁶ + C =

let's distribute and simplify:

{ {[21(x - 1)²] /168} + {[48(x - 1)] /168} + (28/168)} (x - 1)⁶ + C =

{[(x - 1)² /8] + {[2(x - 1)] /7} + (1/6)} (x - 1)⁶ + C =

(expanding)

[(x - 1)⁸ /8] + {[2(x - 1)⁷] /7} + [(x - 1)⁶ /6] + C

I hope it's helpful