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Question: Evaluate the following integral: ∫(x^3 + 2x^2 + 3x + 4) / (x^4 + 4x^3 + 6x^2 + 4x + 1) dx solve it

Question

Question:

Evaluate the following integral:
x3+2x2+3x+4x4+4x3+6x2+4x+1dx \int \frac{x^3 + 2x^2 + 3x + 4}{x^4 + 4x^3 + 6x^2 + 4x + 1} \, dx

solve it

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Solution

1. Break Down the Problem

We need to evaluate the integral

I=x3+2x2+3x+4x4+4x3+6x2+4x+1dx I = \int \frac{x^3 + 2x^2 + 3x + 4}{x^4 + 4x^3 + 6x^2 + 4x + 1} \, dx

2. Relevant Concepts

To solve this integral, we might start by examining the denominator. We see that the polynomial x4+4x3+6x2+4x+1x^4 + 4x^3 + 6x^2 + 4x + 1 could be related to a square of a quadratic polynomial.

3. Analysis and Detail

Let's first rewrite the denominator:

x4+4x3+6x2+4x+1=(x2+2x+1)2=(x+1)4 x^4 + 4x^3 + 6x^2 + 4x + 1 = (x^2 + 2x + 1)^2 = (x + 1)^4

Thus, the integral simplifies to:

I=x3+2x2+3x+4(x+1)4dx I = \int \frac{x^3 + 2x^2 + 3x + 4}{(x + 1)^4} \, dx

Next, we can perform polynomial long division to separate the integral into simpler parts if needed. Start by rewriting the numerator in terms of the denominator:

  1. Divide x3+2x2+3x+4x^3 + 2x^2 + 3x + 4 by x+1x + 1:

x3+2x2+3x+4=(x2+x+2)(x+1)+2 x^3 + 2x^2 + 3x + 4 = (x^2 + x + 2)(x + 1) + 2

Thus, we can rewrite the integral:

I=(x2+x+2(x+1)3+2(x+1)4)dx I = \int \left( \frac{x^2 + x + 2}{(x+1)^3} + \frac{2}{(x+1)^4} \right) \, dx

This gives us two integrals to solve:

I=x2+x+2(x+1)3dx+2(x+1)4dx I = \int \frac{x^2 + x + 2}{(x+1)^3} \, dx + \int \frac{2}{(x+1)^4} \, dx

4. Verify and Summarize

Now we can evaluate each integral separately:

  1. For 2(x+1)4dx\int \frac{2}{(x + 1)^4} \, dx:

2(x+1)4dx=23(x+1)3+C1 \int \frac{2}{(x + 1)^4} \, dx = -\frac{2}{3(x + 1)^3} + C_1

  1. For x2+x+2(x+1)3dx\int \frac{x^2 + x + 2}{(x + 1)^3} \, dx, we can apply the substitution u=x+1 u = x + 1 which simplifies the calculation.

=(u1)2+(u1)+2u3du=u22u+1+u1+2u3du=u2u+2u3du = \int \frac{(u-1)^2 + (u-1) + 2}{u^3} \, du = \int \frac{u^2 - 2u + 1 + u - 1 + 2}{u^3} \, du = \int \frac{u^2 - u + 2}{u^3} \, du

Break it down further:

=(1u1u2+2u3)du = \int \left( \frac{1}{u} - \frac{1}{u^2} + \frac{2}{u^3} \right) \, du

Evaluating this gives:

=lnu+1u22u2+C2=lnx+11(x+1)2+C2 = \ln |u| + \frac{1}{u} - \frac{2}{2u^2} + C_2 = \ln |x + 1| - \frac{1}{(x + 1)^2} + C_2

Combining these results:

I=[lnx+11(x+1)223(x+1)3+C] I = \left[ \ln |x + 1| - \frac{1}{(x + 1)^2} - \frac{2}{3(x + 1)^3} + C \right]

Final Answer

Thus, the evaluated integral is:

I=lnx+11(x+1)223(x+1)3+C I = \ln |x + 1| - \frac{1}{(x + 1)^2} - \frac{2}{3(x + 1)^3} + C where CC is the constant of integration.

This problem has been solved

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