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3.5 Q-26

Question Statement

Evaluate the integral:
∫3x+1(4x2+1)(x2βˆ’x+1),dx\int \frac{3 x+1}{\left(4 x^{2}+1\right)\left(x^{2}-x+1\right)} , dx


Background and Explanation

This problem involves a rational function that requires partial fraction decomposition. The key steps involve expressing the given rational function as the sum of simpler fractions and then integrating each term separately. We will break the denominator into two parts, each of which will be handled using basic integration techniques.


Solution

We begin by expressing the integrand as the sum of two simpler fractions: 3x+1(4x2+1)(x2βˆ’x+1)=Ax+B4x2+1+Cx+Dx2βˆ’x+1\frac{3x+1}{\left(4x^{2}+1\right)\left(x^{2}-x+1\right)} = \frac{Ax + B}{4x^2 + 1} + \frac{Cx + D}{x^2 - x + 1}

Step 1: Multiply both sides by the denominator

To eliminate the denominators, multiply both sides of the equation by (4x2+1)(x2βˆ’x+1)(4x^2 + 1)(x^2 - x + 1): 3x+1=(Ax+B)(x2βˆ’x+1)+(Cx+D)(4x2+1)3x + 1 = (Ax + B)(x^2 - x + 1) + (Cx + D)(4x^2 + 1)

Step 2: Expand both sides

Expanding the terms on the right-hand side:

3x+1=A(x3βˆ’x2+x)+B(x2βˆ’x+1)+C(4x2+x)+D(4x2+1)=Ax3βˆ’Ax2+Ax+Bx2βˆ’Bx+B+4Cx2+Cx+4Dx2+D\begin{aligned} 3x + 1 &= A(x^3 - x^2 + x) + B(x^2 - x + 1) + C(4x^2 + x) + D(4x^2 + 1) &= A x^3 - A x^2 + A x + B x^2 - B x + B + 4 C x^2 + C x + 4 D x^2 + D \end{aligned}

Step 3: Group terms by powers of xx

Group all the powers of xx:

=Ax3+(βˆ’A+B+4C+4D)x2+(Aβˆ’B+C)x+(B+D)\begin{aligned} &= A x^3 + (-A + B + 4C + 4D) x^2 + (A - B + C) x + (B + D) \end{aligned}

Step 4: Compare coefficients

Now, equate the coefficients of like powers of xx on both sides:

  • Coefficient of x3x^3: 0=A+4C0 = A + 4C
  • Coefficient of x2x^2: 0=βˆ’A+B+4D0 = -A + B + 4D
  • Coefficient of xx: 3=Aβˆ’B+C3 = A - B + C
  • Constant term: 1=B+D1 = B + D

Step 5: Solve the system of equations

From A+4C=0A + 4C = 0, we get:

A=βˆ’4CA = -4C

Substitute into Aβˆ’B+C=3A - B + C = 3:

βˆ’4Cβˆ’B+C=3β‡’βˆ’3Cβˆ’B=3(EquationΒ 1)-4C - B + C = 3 \quad \Rightarrow \quad -3C - B = 3 \quad \text{(Equation 1)}

From B+D=1B + D = 1, we get:

B=1βˆ’DB = 1 - D

Substitute into βˆ’A+B+4D=0-A + B + 4D = 0:

βˆ’(βˆ’4C)+(1βˆ’D)+4D=0β‡’4C+1βˆ’D+4D=0β‡’4C+1+3D=0(EquationΒ 2)-(-4C) + (1 - D) + 4D = 0 \quad \Rightarrow \quad 4C + 1 - D + 4D = 0 \quad \Rightarrow \quad 4C + 1 + 3D = 0 \quad \text{(Equation 2)}

Now, we solve these equations. First, substitute A=βˆ’4CA = -4C and B=1βˆ’DB = 1 - D into the equations and solve for CC, BB, and DD.

From Equation 2, solve for CC:

4C=βˆ’1βˆ’3Dβ‡’C=βˆ’1+3D44C = -1 - 3D \quad \Rightarrow \quad C = -\frac{1 + 3D}{4}

Substitute this into Equation 1:

βˆ’3(βˆ’1+3D4)βˆ’(1βˆ’D)=3β‡’3(1+3D)4βˆ’1+D=3-3\left(-\frac{1 + 3D}{4}\right) - (1 - D) = 3 \quad \Rightarrow \quad \frac{3(1 + 3D)}{4} - 1 + D = 3

Multiply through by 4:

3(1+3D)βˆ’4(1βˆ’D)=12β‡’3+9Dβˆ’4+4D=12β‡’13D=133(1 + 3D) - 4(1 - D) = 12 \quad \Rightarrow \quad 3 + 9D - 4 + 4D = 12 \quad \Rightarrow \quad 13D = 13

Thus, D=1D = 1.

Now, substitute D=1D = 1 into B=1βˆ’DB = 1 - D and C=βˆ’1+3D4C = -\frac{1 + 3D}{4}:

B=0,C=βˆ’1B = 0, \quad C = -1

Finally, substitute C=βˆ’1C = -1 into A=βˆ’4CA = -4C:

A=4A = 4

Step 6: Substitute back into the partial fractions

Now we can write the partial fraction decomposition as:

3x+1(4x2+1)(x2βˆ’x+1)=4x4x2+1βˆ’1x2βˆ’x+1\frac{3x+1}{(4x^2 + 1)(x^2 - x + 1)} = \frac{4x}{4x^2 + 1} - \frac{1}{x^2 - x + 1}

Step 7: Integrate each term

We now integrate each term separately:

  1. For ∫4x4x2+1,dx\int \frac{4x}{4x^2 + 1} , dx, use the substitution u=4x2+1u = 4x^2 + 1, so du=8x,dxdu = 8x , dx. This results in:
12∫uβˆ’1,du=12ln⁑∣u∣=12ln⁑(4x2+1)\frac{1}{2} \int u^{-1} , du = \frac{1}{2} \ln |u| = \frac{1}{2} \ln (4x^2 + 1)
  1. For ∫1x2βˆ’x+1,dx\int \frac{1}{x^2 - x + 1} , dx, complete the square in the denominator and use the formula for the integral of 1x2+a2\frac{1}{x^2 + a^2}:
x2βˆ’x+1=(xβˆ’12)2+34x^2 - x + 1 = \left( x - \frac{1}{2} \right)^2 + \frac{3}{4}

Thus,

∫1x2βˆ’x+1,dx=23tanβ‘βˆ’1(2xβˆ’13)\int \frac{1}{x^2 - x + 1} , dx = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x - 1}{\sqrt{3}} \right)

Step 8: Combine results

Combining the results, we get the final answer:

∫3x+1(4x2+1)(x2βˆ’x+1),dx=12ln⁑(4x2+1x2βˆ’x+1)+13tanβ‘βˆ’1(2xβˆ’13)+C\int \frac{3x + 1}{(4x^2 + 1)(x^2 - x + 1)} , dx = \frac{1}{2} \ln \left( \frac{4x^2 + 1}{x^2 - x + 1} \right) + \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{2x - 1}{\sqrt{3}} \right) + C

Key Formulas or Methods Used

  • Partial Fraction Decomposition: To break down complex rational expressions into simpler fractions.
  • Substitution: Used for simplifying integrals.
  • Standard Integrals: Specifically for terms like ∫1x2+a2,dx\int \frac{1}{x^2 + a^2} , dx and the logarithmic integration for ∫4x4x2+1,dx\int \frac{4x}{4x^2 + 1} , dx.

Summary of Steps

  1. Set up the partial fraction decomposition.
  2. Multiply both sides by the common denominator and expand.
  3. Equate the coefficients of powers of xx.
  4. Solve the system of equations for AA, BB, CC, and DD.
  5. Substitute the values back into the partial fractions.
  6. Integrate each term using known integration formulas.
  7. Combine the results to obtain the final answer.