Integrate 12x^2(3+2x)^5

Question

using Substitution Method?

I got almost the right answer but not quite. Show all lines of working please.

Answer 1

∫ 12x^2(3+2x)^5 dx =
∫ (12x^2+18x)(3+2x)^5 dx - ∫ (18x+27)(3+2x)^5 dx + 27∫(3+2x)^5 dx =
6∫ x(3+2x)^6 dx - 9∫ (3+2x)^6 dx + 27∫(3+2x)^5 dx =
3∫ (3+2x)(3+2x)^6 dx - 18∫ (3+2x)^6 dx + 27∫(3+2x)^5 dx =
(3/2)∫(3+2x)^7 d(3+2x) - 9∫ (3+2x)^6 d(3+2x) + (27/2)∫(3+2x)^5 d(3+2x) =
(3/16)(3+2x)^8 - (9/7) (3+2x)^7 + (9/2)(3+2x)^6+C

Answer 2

Integrate 12x²(3+2x)⁵

Take (3+2x)⁵ = (3 + 2x)²(3 + 2x)³

Principal algebraic expressions and formulas: (a+b)² = a²+2ab+b² and (a+b)³=a³+3a²b+3ab²+b³

= (9 + 4x² + 12x)(27 + 8x³ + 54x + 18x )

FOIL method: the product of two binomials is the sum of the products of the First terms, the Outer terms, the Inner terms and the Last terms.

= 32x⁵+ 240x⁴ + 720x³ + 1080x² + 810x + 243.

Integrate 12x²(3+2x)⁵dx

ʃ12x²(32x⁵+ 240x⁴ + 720x³ + 1080x² + 810x + 243)dx

12ʃ(32x⁷+ 240x⁶ + 720x⁵ + 1080x⁴ + 810x³ + 243x²)dx

12[ʃ(32x⁷)dx + 240ʃ(x⁶)dx + 720ʃ(x⁵)dx + 1080ʃ(x⁴)dx + 810ʃ(x³)dx + 243ʃ(x²)dx]

Integrals of rational functions: ʃxⁿdx = [xⁿ⁺¹/n+1]+C

12[(32x⁸/8) + 240(x⁷/7) + 720(x⁶/6) + 1080(x⁵/5) + 810(x⁴/4) + 243(x³/3)] + C

12[4x⁸) + (240/7)x⁷ + 120x⁶ + 216x⁵ + (405/2)x⁴ + 81x³] + C

Integrate 12x²(3+2x)⁵dx is 12[4x⁸) + (240/7)x⁷ + 120x⁶ + 216x⁵ + (405/2)x⁴ + 81x³] + C.

Answer 3

Integration by substitution method :

ʃ 12x²(3+2x)⁵ dx.

Let, u = 3 + 2x → x = (u - 3)/2.

du = 2 dx.

ʃ 12x²(3+2x)⁵ dx

= ʃ 6[(u - 3)/2]²u⁵ du

= 6/4ʃ [u - 3]²u⁵ du

= 3/2ʃ [[u² - 6u + 9]u⁵] du

= 3/2ʃ [u⁷ - 6u⁶ + 9u⁵] du

Recall the formula : ʃxⁿdx = [xⁿ⁺¹/n+1]+C

= 3/2[(u⁸/8) - 6(u⁷/7) + 9(u⁶/6)] + C

= (3/16)u⁸ - (9/7)u⁷ + (9/4)u⁶ + C.

Put, u = 3 + 2x.

= (3/16)(3 + 2x)⁸ - (9/7)(3 + 2x)⁷ + (9/4)(3 + 2x)⁶ + C.

Therefore, ʃ 12x²(3+2x)⁵ dx = (3/16)(3 + 2x)⁸ - (9/7)(3 + 2x)⁷ + (9/4)(3 + 2x)⁶ + C.