Question Statement
Evaluate the integral:
β« 3 x + 1 ( 4 x 2 + 1 ) ( x 2 β x + 1 ) , d x \int \frac{3 x+1}{\left(4 x^{2}+1\right)\left(x^{2}-x+1\right)} , dx β« ( 4 x 2 + 1 ) ( x 2 β x + 1 ) 3 x + 1 β , d x
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:
3 x + 1 ( 4 x 2 + 1 ) ( x 2 β x + 1 ) = A x + B 4 x 2 + 1 + C x + D x 2 β 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} ( 4 x 2 + 1 ) ( x 2 β x + 1 ) 3 x + 1 β = 4 x 2 + 1 A x + B β + x 2 β x + 1 C x + D β
Step 1: Multiply both sides by the denominator
To eliminate the denominators, multiply both sides of the equation by ( 4 x 2 + 1 ) ( x 2 β x + 1 ) (4x^2 + 1)(x^2 - x + 1) ( 4 x 2 + 1 ) ( x 2 β x + 1 ) :
3 x + 1 = ( A x + B ) ( x 2 β x + 1 ) + ( C x + D ) ( 4 x 2 + 1 ) 3x + 1 = (Ax + B)(x^2 - x + 1) + (Cx + D)(4x^2 + 1) 3 x + 1 = ( A x + B ) ( x 2 β x + 1 ) + ( C x + D ) ( 4 x 2 + 1 )
Step 2: Expand both sides
Expanding the terms on the right-hand side:
3 x + 1 = A ( x 3 β x 2 + x ) + B ( x 2 β x + 1 ) + C ( 4 x 2 + x ) + D ( 4 x 2 + 1 ) = A x 3 β A x 2 + A x + B x 2 β B x + B + 4 C x 2 + C x + 4 D x 2 + 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} 3 x + 1 β = A ( x 3 β x 2 + x ) + B ( x 2 β x + 1 ) + C ( 4 x 2 + x ) + D ( 4 x 2 + 1 ) β = A x 3 β A x 2 + A x + B x 2 β B x + B + 4 C x 2 + C x + 4 D x 2 + D β
Step 3: Group terms by powers of x x x
Group all the powers of x x x :
= A x 3 + ( β A + B + 4 C + 4 D ) x 2 + ( A β B + C ) x + ( B + D ) \begin{aligned}
&= A x^3 + (-A + B + 4C + 4D) x^2 + (A - B + C) x + (B + D)
\end{aligned} β = A x 3 + ( β A + B + 4 C + 4 D ) x 2 + ( A β B + C ) x + ( B + D ) β
Step 4: Compare coefficients
Now, equate the coefficients of like powers of x x x on both sides:
Coefficient of x 3 x^3 x 3 : 0 = A + 4 C 0 = A + 4C 0 = A + 4 C
Coefficient of x 2 x^2 x 2 : 0 = β A + B + 4 D 0 = -A + B + 4D 0 = β A + B + 4 D
Coefficient of x x x : 3 = A β B + C 3 = A - B + C 3 = A β B + C
Constant term: 1 = B + D 1 = B + D 1 = B + D
Step 5: Solve the system of equations
From A + 4 C = 0 A + 4C = 0 A + 4 C = 0 , we get:
A = β 4 C A = -4C A = β 4 C
Substitute into A β B + C = 3 A - B + C = 3 A β B + C = 3 :
β 4 C β B + C = 3 β β 3 C β B = 3 (EquationΒ 1) -4C - B + C = 3 \quad \Rightarrow \quad -3C - B = 3 \quad \text{(Equation 1)} β 4 C β B + C = 3 β β 3 C β B = 3 (EquationΒ 1)
From B + D = 1 B + D = 1 B + D = 1 , we get:
B = 1 β D B = 1 - D B = 1 β D
Substitute into β A + B + 4 D = 0 -A + B + 4D = 0 β A + B + 4 D = 0 :
β ( β 4 C ) + ( 1 β D ) + 4 D = 0 β 4 C + 1 β D + 4 D = 0 β 4 C + 1 + 3 D = 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)} β ( β 4 C ) + ( 1 β D ) + 4 D = 0 β 4 C + 1 β D + 4 D = 0 β 4 C + 1 + 3 D = 0 (EquationΒ 2)
Now, we solve these equations. First, substitute A = β 4 C A = -4C A = β 4 C and B = 1 β D B = 1 - D B = 1 β D into the equations and solve for C C C , B B B , and D D D .
From Equation 2, solve for C C C :
4 C = β 1 β 3 D β C = β 1 + 3 D 4 4C = -1 - 3D \quad \Rightarrow \quad C = -\frac{1 + 3D}{4} 4 C = β 1 β 3 D β C = β 4 1 + 3 D β
Substitute this into Equation 1:
β 3 ( β 1 + 3 D 4 ) β ( 1 β D ) = 3 β 3 ( 1 + 3 D ) 4 β 1 + D = 3 -3\left(-\frac{1 + 3D}{4}\right) - (1 - D) = 3 \quad \Rightarrow \quad \frac{3(1 + 3D)}{4} - 1 + D = 3 β 3 ( β 4 1 + 3 D β ) β ( 1 β D ) = 3 β 4 3 ( 1 + 3 D ) β β 1 + D = 3
Multiply through by 4:
3 ( 1 + 3 D ) β 4 ( 1 β D ) = 12 β 3 + 9 D β 4 + 4 D = 12 β 13 D = 13 3(1 + 3D) - 4(1 - D) = 12 \quad \Rightarrow \quad 3 + 9D - 4 + 4D = 12 \quad \Rightarrow \quad 13D = 13 3 ( 1 + 3 D ) β 4 ( 1 β D ) = 12 β 3 + 9 D β 4 + 4 D = 12 β 13 D = 13
Thus, D = 1 D = 1 D = 1 .
Now, substitute D = 1 D = 1 D = 1 into B = 1 β D B = 1 - D B = 1 β D and C = β 1 + 3 D 4 C = -\frac{1 + 3D}{4} C = β 4 1 + 3 D β :
B = 0 , C = β 1 B = 0, \quad C = -1 B = 0 , C = β 1
Finally, substitute C = β 1 C = -1 C = β 1 into A = β 4 C A = -4C A = β 4 C :
A = 4 A = 4 A = 4
Step 6: Substitute back into the partial fractions
Now we can write the partial fraction decomposition as:
3 x + 1 ( 4 x 2 + 1 ) ( x 2 β x + 1 ) = 4 x 4 x 2 + 1 β 1 x 2 β x + 1 \frac{3x+1}{(4x^2 + 1)(x^2 - x + 1)} = \frac{4x}{4x^2 + 1} - \frac{1}{x^2 - x + 1} ( 4 x 2 + 1 ) ( x 2 β x + 1 ) 3 x + 1 β = 4 x 2 + 1 4 x β β x 2 β x + 1 1 β
Step 7: Integrate each term
We now integrate each term separately:
For β« 4 x 4 x 2 + 1 , d x \int \frac{4x}{4x^2 + 1} , dx β« 4 x 2 + 1 4 x β , d x , use the substitution u = 4 x 2 + 1 u = 4x^2 + 1 u = 4 x 2 + 1 , so d u = 8 x , d x du = 8x , dx d u = 8 x , d x . This results in:
1 2 β« u β 1 , d u = 1 2 ln β‘ β£ u β£ = 1 2 ln β‘ ( 4 x 2 + 1 ) \frac{1}{2} \int u^{-1} , du = \frac{1}{2} \ln |u| = \frac{1}{2} \ln (4x^2 + 1) 2 1 β β« u β 1 , d u = 2 1 β ln β£ u β£ = 2 1 β ln ( 4 x 2 + 1 )
For β« 1 x 2 β x + 1 , d x \int \frac{1}{x^2 - x + 1} , dx β« x 2 β x + 1 1 β , d x , complete the square in the denominator and use the formula for the integral of 1 x 2 + a 2 \frac{1}{x^2 + a^2} x 2 + a 2 1 β :
x 2 β x + 1 = ( x β 1 2 ) 2 + 3 4 x^2 - x + 1 = \left( x - \frac{1}{2} \right)^2 + \frac{3}{4} x 2 β x + 1 = ( x β 2 1 β ) 2 + 4 3 β
Thus,
β« 1 x 2 β x + 1 , d x = 2 3 tan β‘ β 1 ( 2 x β 1 3 ) \int \frac{1}{x^2 - x + 1} , dx = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x - 1}{\sqrt{3}} \right) β« x 2 β x + 1 1 β , d x = 3 β 2 β tan β 1 ( 3 β 2 x β 1 β )
Step 8: Combine results
Combining the results, we get the final answer:
β« 3 x + 1 ( 4 x 2 + 1 ) ( x 2 β x + 1 ) , d x = 1 2 ln β‘ ( 4 x 2 + 1 x 2 β x + 1 ) + 1 3 tan β‘ β 1 ( 2 x β 1 3 ) + 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 β« ( 4 x 2 + 1 ) ( x 2 β x + 1 ) 3 x + 1 β , d x = 2 1 β ln ( x 2 β x + 1 4 x 2 + 1 β ) + 3 β 1 β tan β 1 ( 3 β 2 x β 1 β ) + C
Partial Fraction Decomposition : To break down complex rational expressions into simpler fractions.
Substitution : Used for simplifying integrals.
Standard Integrals : Specifically for terms like β« 1 x 2 + a 2 , d x \int \frac{1}{x^2 + a^2} , dx β« x 2 + a 2 1 β , d x and the logarithmic integration for β« 4 x 4 x 2 + 1 , d x \int \frac{4x}{4x^2 + 1} , dx β« 4 x 2 + 1 4 x β , d x .
Summary of Steps
Set up the partial fraction decomposition.
Multiply both sides by the common denominator and expand.
Equate the coefficients of powers of x x x .
Solve the system of equations for A A A , B B B , C C C , and D D D .
Substitute the values back into the partial fractions.
Integrate each term using known integration formulas.
Combine the results to obtain the final answer.