Math Problem Statement

∫ (x 2 +x+1) 2

dx ​

Solution

To solve the integral (x2+x+1)2dx\int (x^2 + x + 1)^2 \, dx, we can follow these steps:

Step 1: Expand the integrand

First, expand the square of the polynomial: (x2+x+1)2=(x2+x+1)(x2+x+1)(x^2 + x + 1)^2 = (x^2 + x + 1)(x^2 + x + 1) Expanding this, we get: (x2+x+1)(x2+x+1)=x4+2x3+3x2+2x+1(x^2 + x + 1)(x^2 + x + 1) = x^4 + 2x^3 + 3x^2 + 2x + 1

Step 2: Integrate term by term

Now, integrate each term in the expanded polynomial separately: (x4+2x3+3x2+2x+1)dx\int (x^4 + 2x^3 + 3x^2 + 2x + 1) \, dx This can be done as follows: x4dx+2x3dx+3x2dx+2xdx+1dx\int x^4 \, dx + 2\int x^3 \, dx + 3\int x^2 \, dx + 2\int x \, dx + \int 1 \, dx

Step 3: Compute each integral

  1. x4dx=x55\int x^4 \, dx = \frac{x^5}{5}
  2. 2x3dx=2(x44)=x422\int x^3 \, dx = 2\left(\frac{x^4}{4}\right) = \frac{x^4}{2}
  3. 3x2dx=3(x33)=x33\int x^2 \, dx = 3\left(\frac{x^3}{3}\right) = x^3
  4. 2xdx=2(x22)=x22\int x \, dx = 2\left(\frac{x^2}{2}\right) = x^2
  5. 1dx=x\int 1 \, dx = x

Step 4: Combine the results

Combine all these results to get the final answer: x55+x42+x3+x2+x+C\frac{x^5}{5} + \frac{x^4}{2} + x^3 + x^2 + x + C where CC is the constant of integration.

So, the integral is: (x2+x+1)2dx=x55+x42+x3+x2+x+C\int (x^2 + x + 1)^2 \, dx = \frac{x^5}{5} + \frac{x^4}{2} + x^3 + x^2 + x + C

Would you like more details on any specific step or have any questions?

Relative Questions

  1. How do you expand polynomials using the distributive property?
  2. What is the power rule for integration, and how is it applied?
  3. Why is the constant of integration important in indefinite integrals?
  4. How do you verify the result of an integration problem?
  5. Can you apply the method of integration by parts to solve similar integrals?

Tip

When integrating polynomials, always remember to expand the expression first if it simplifies the integration process.

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Math Problem Analysis

Mathematical Concepts

Integration
Polynomials

Formulas

Power rule for integration

Theorems

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Suitable Grade Level

Advanced High School / College