Question Statement Evaluate the following integral:
∫ 2 x 2 − 2 x 4 + x 2 + 1 , d x \int \frac{2 x^{2}-2}{x^{4} + x^{2} + 1} , dx ∫ x 4 + x 2 + 1 2 x 2 − 2 , d x Background and Explanation This problem involves integrating a rational function where the denominator is a quartic expression. A typical approach for such integrals is partial fraction decomposition , which breaks down the rational function into simpler fractions that are easier to integrate. To apply this technique, we first need to factor the denominator and express the function as a sum of simpler fractions.
Solution Step 1: Factor the Denominator
The denominator x 4 + x 2 + 1 x^{4} + x^{2} + 1 x 4 + x 2 + 1
x 4 + x 2 + 1 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) x^{4} + x^{2} + 1 = \left(x^{2} + x + 1\right)\left(x^{2} - x + 1\right) x 4 + x 2 + 1 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) We use this factorization to express the given rational function in a more manageable form.
Step 2: Partial Fraction Decomposition
We aim to express the integrand as the sum of two simpler fractions:
2 x 2 − 2 ( x 2 + x + 1 ) ( x 2 − x + 1 ) = A x + B x 2 + x + 1 + C x + D x 2 − x + 1 \frac{2x^{2} - 2}{(x^{2} + x + 1)(x^{2} - x + 1)} = \frac{A x + B}{x^{2} + x + 1} + \frac{C x + D}{x^{2} - x + 1} ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 x 2 − 2 = x 2 + x + 1 A x + B + x 2 − x + 1 C x + D Multiplying both sides by the denominator ( x 2 + x + 1 ) ( x 2 − x + 1 ) (x^{2} + x + 1)(x^{2} - x + 1) ( x 2 + x + 1 ) ( x 2 − x + 1 )
2 x 2 − 2 = ( A x + B ) ( x 2 − x + 1 ) + ( C x + D ) ( x 2 + x + 1 ) 2 x^{2} - 2 = (A x + B)(x^{2} - x + 1) + (C x + D)(x^{2} + x + 1) 2 x 2 − 2 = ( A x + B ) ( x 2 − x + 1 ) + ( C x + D ) ( x 2 + x + 1 ) Step 3: Expand and Simplify
Expand both terms on the right-hand side:
2 x 2 − 2 = ( A x + B ) ( x 2 − x + 1 ) + ( C x + D ) ( x 2 + x + 1 ) 2 x^{2} - 2 = (A x + B)(x^{2} - x + 1) + (C x + D)(x^{2} + x + 1) 2 x 2 − 2 = ( A x + B ) ( x 2 − x + 1 ) + ( C x + D ) ( x 2 + x + 1 ) Expanding each term:
( A x + B ) ( x 2 − x + 1 ) = A x 3 − A x 2 + A x + B x 2 − B x + B (A x + B)(x^{2} - x + 1) = A x^{3} - A x^{2} + A x + B x^{2} - B x + B ( A x + B ) ( x 2 − x + 1 ) = A x 3 − A x 2 + A x + B x 2 − B x + B ( C x + D ) ( x 2 + x + 1 ) = C x 3 + C x 2 + C x + D x 2 + D x + D (C x + D)(x^{2} + x + 1) = C x^{3} + C x^{2} + C x + D x^{2} + D x + D ( C x + D ) ( x 2 + x + 1 ) = C x 3 + C x 2 + C x + D x 2 + D x + D Simplifying the equation:
2 x 2 − 2 = ( A + C ) x 3 + ( − A + B + C + D ) x 2 + ( A − B + C + D ) x + ( B + D ) 2 x^{2} - 2 = (A + C) x^{3} + (-A + B + C + D) x^{2} + (A - B + C + D) x + (B + D) 2 x 2 − 2 = ( A + C ) x 3 + ( − A + B + C + D ) x 2 + ( A − B + C + D ) x + ( B + D ) Step 4: Equate Coefficients
Now, equate the coefficients of corresponding powers of x x x
Coefficient of x 3 x^{3} x 3 A + C = 0 A + C = 0 A + C = 0
Coefficient of x 2 x^{2} x 2 − A + B + C + D = 2 -A + B + C + D = 2 − A + B + C + D = 2
Coefficient of x x x A − B + C + D = 0 A - B + C + D = 0 A − B + C + D = 0
Constant term: B + D = − 2 B + D = -2 B + D = − 2
Step 5: Solve the System of Equations
From A + C = 0 A + C = 0 A + C = 0 C = − A C = -A C = − A
From B + D = − 2 B + D = -2 B + D = − 2 D = − 2 − B D = -2 - B D = − 2 − B
Substituting C = − A C = -A C = − A D = − 2 − B D = -2 - B D = − 2 − B A − B + C + D = 0 A - B + C + D = 0 A − B + C + D = 0
A − B − A − 2 − B = 0 ⇒ − 2 B = 2 ⇒ B = − 1 A - B - A - 2 - B = 0 \quad \Rightarrow \quad -2B = 2 \quad \Rightarrow \quad B = -1 A − B − A − 2 − B = 0 ⇒ − 2 B = 2 ⇒ B = − 1 Substitute B = − 1 B = -1 B = − 1 B + D = − 2 B + D = -2 B + D = − 2
− 1 + D = − 2 ⇒ D = − 1 -1 + D = -2 \quad \Rightarrow \quad D = -1 − 1 + D = − 2 ⇒ D = − 1 Now, substitute B = − 1 B = -1 B = − 1 D = − 1 D = -1 D = − 1 − A + B + C + D = 2 -A + B + C + D = 2 − A + B + C + D = 2
− A − 1 − A − 1 = 2 ⇒ − 2 A − 2 = 2 ⇒ A = − 2 -A - 1 - A - 1 = 2 \quad \Rightarrow \quad -2A - 2 = 2 \quad \Rightarrow \quad A = -2 − A − 1 − A − 1 = 2 ⇒ − 2 A − 2 = 2 ⇒ A = − 2 Thus, C = − A = 2 C = -A = 2 C = − A = 2
Step 6: Write the Partial Fractions
Now that we have the values of A A A B B B C C C D D D
2 x 2 − 2 ( x 2 + x + 1 ) ( x 2 − x + 1 ) = − 2 x − 1 x 2 + x + 1 + 2 x − 1 x 2 − x + 1 \frac{2 x^{2} - 2}{(x^{2} + x + 1)(x^{2} - x + 1)} = \frac{-2x - 1}{x^{2} + x + 1} + \frac{2x - 1}{x^{2} - x + 1} ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 x 2 − 2 = x 2 + x + 1 − 2 x − 1 + x 2 − x + 1 2 x − 1 Step 7: Integrate the Terms
Now, we can integrate each term separately:
∫ − 2 x − 1 x 2 + x + 1 , d x and ∫ 2 x − 1 x 2 − x + 1 , d x \int \frac{-2x - 1}{x^{2} + x + 1} , dx \quad \text{and} \quad \int \frac{2x - 1}{x^{2} - x + 1} , dx ∫ x 2 + x + 1 − 2 x − 1 , d x and ∫ x 2 − x + 1 2 x − 1 , d x Both integrals are standard, resulting in the following:
− ln ∣ x 2 + x + 1 ∣ + ln ∣ x 2 − x + 1 ∣ + C -\ln |x^{2} + x + 1| + \ln |x^{2} - x + 1| + C − ln ∣ x 2 + x + 1∣ + ln ∣ x 2 − x + 1∣ + C Thus, the final answer is:
ln ∣ x 2 − x + 1 x 2 + x + 1 ∣ + C \ln \left|\frac{x^{2} - x + 1}{x^{2} + x + 1}\right| + C ln x 2 + x + 1 x 2 − x + 1 + C
Partial Fraction Decomposition : This method breaks down the given rational function into simpler fractions.Standard Integral Forms : Used for logarithmic integration of quadratic expressions.
Summary of Steps
Factor the denominator x 4 + x 2 + 1 x^{4} + x^{2} + 1 x 4 + x 2 + 1 ( x 2 + x + 1 ) ( x 2 − x + 1 ) (x^{2} + x + 1)(x^{2} - x + 1) ( x 2 + x + 1 ) ( x 2 − x + 1 ) Set up the partial fraction decomposition with unknown coefficients A A A B B B C C C D D D Expand the terms on the right-hand side and equate the coefficients of powers of x x x Solve the system of equations for A A A B B B C C C D D D Rewrite the integrand using the values of the coefficients.Integrate each term , using the standard logarithmic integration.Simplify to get the final answer:
ln ∣ x 2 − x + 1 x 2 + x + 1 ∣ + C \ln \left|\frac{x^{2} - x + 1}{x^{2} + x + 1}\right| + C ln x 2 + x + 1 x 2 − x + 1 + C 
