. Math Help Please .

Could someone answer this question so I understand it. Thanks

Find the indefinite integral

∫ p(p + 3)^5 dp

Use C as the arbitrary constant

p∫(p + 3)^5

let U = (p+3)

so du = 1dp

p ∫(U)^5 du = p[ (1/6)(U)^6] + C

1/6p(U)^6 + C

1/6p(p+3)^6 + C

This is absolutely crazy, because the answer that this person gave is wrong.

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∫p(p + 3)⁵dp

(p + 3)⁵ = p⁵ + 5×3p⁴ + 10×9p³ + 10×27p² + 5×81p + 243

p(p + 3)⁵ = p⁶ + 15p⁵ + 90p⁴ + 270p³ + 405p² + 243p

Therefore,

∫p(p + 3)⁵dp = ∫(p⁶ + 15p⁵ + 90p⁴ + 270p³ + 405p² + 243p)dp

= p⁷/7 + 15p⁶/6 + 90p⁵/5 + 270p⁴/4 + 405p³/3 + 243p²/2 + C

= p⁷/7 + 5p⁶/2 + 18p⁵ + 135p⁴/2 + 135p³ + 243p²/2 + C

I will let you simplify this further if you wish.

ProfRay

Integrate the original integrand by substitution:
∫ p(p + 3)⁵ dp
Let u = p + 3,
du / dp = 1
du = dp
p = u - 3
∫ (u - 3)u⁵ du = ∫ u⁶ du - 3 ∫ u⁵ du
∫ (u - 3)u⁵ du = u⁷ / 7 - u⁶ / 2 + C
Since u = p + 3,
∫ p(p + 3)⁵ dp = (p + 3)⁷ / 7 - (p + 3)⁶ / 2 + C