Math Problem Statement

∫ -(12x + 4)(4 - (3x^2 + 2x))^5 dx

Solution

The given integral is:

(12x+4)(4(3x2+2x))5dx\int - (12x + 4)(4 - (3x^2 + 2x))^5 \, dx

Steps to solve the integral:

  1. Substitution: Let: u=4(3x2+2x)u = 4 - (3x^2 + 2x) Then, differentiate uu with respect to xx: dudx=(6x+2)\frac{du}{dx} = -(6x + 2) Hence, du=(6x+2)dxdu = -(6x + 2) \, dx Notice that the factor (12x+4)(12x + 4) can be factored as: 12x+4=2(6x+2)12x + 4 = 2(6x + 2) This matches the derivative of uu, which simplifies the integral.

  2. Substitute into the integral: Rewriting the integral in terms of uu: (12x+4)(4(3x2+2x))5dx=2(6x+2)u5dx\int - (12x + 4)(4 - (3x^2 + 2x))^5 \, dx = \int - 2(6x + 2) u^5 \, dx Since du=(6x+2)dxdu = -(6x + 2) dx, we have: 2(6x+2)u5dx=2u5du\int -2(6x + 2) u^5 \, dx = \int 2 u^5 \, du

  3. Integrate: Now, integrate with respect to uu: 2u5du=26+1u6=27u6+C\int 2 u^5 \, du = \frac{2}{6+1} u^{6} = \frac{2}{7} u^6 + C

  4. Substitute back u=4(3x2+2x)u = 4 - (3x^2 + 2x): Finally, substitute back the expression for uu: 27(4(3x2+2x))6+C\frac{2}{7} \left(4 - (3x^2 + 2x)\right)^6 + C

Thus, the solution to the integral is:

27(4(3x2+2x))6+C\frac{2}{7} \left(4 - (3x^2 + 2x)\right)^6 + C


Would you like more details on any step, or do you have any further questions?

Here are some related questions to consider:

  1. What is the significance of substitution in integration?
  2. How would the result change if the exponent inside the integral was different?
  3. Can integration by parts be used here instead of substitution?
  4. How would the integration process change if (12x+4)(12x + 4) had a different form?
  5. What are some common mistakes to watch for when applying substitution in integrals?

Tip: Always ensure that the differential dudu is fully accounted for when using substitution, as missing or incorrect terms can lead to errors.

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Math Problem Analysis

Mathematical Concepts

Calculus
Integration
Substitution Method

Formulas

∫ u^n du = (1/(n+1)) u^(n+1) + C

Theorems

Substitution Rule in Integration

Suitable Grade Level

College level or Advanced High School