# using Substitution Method?

I got almost the right answer but not quite. Show all lines of working please.
asked Feb 25, 2013 in CALCULUS

∫ 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
+1 vote

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.

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.