Question Statement Evaluate the following definite integral:
∫ 2 3 3 x 2 − 2 x + 1 ( x − 1 ) ( x 2 + 1 ) , d x \int_2^3 \frac{3x^2 - 2x + 1}{(x - 1)(x^2 + 1)} , dx ∫ 2 3 ( x − 1 ) ( x 2 + 1 ) 3 x 2 − 2 x + 1 , d x Background and Explanation This problem requires using partial fraction decomposition to break down a complicated rational function into simpler parts that are easier to integrate. The general process involves expressing the integrand as a sum of simpler fractions, finding the values of the constants, and then integrating each term separately.
Solution Step 1: Set up the partial fraction decomposition
We want to express the integrand:
3 x 2 − 2 x + 1 ( x − 1 ) ( x 2 + 1 ) \frac{3x^2 - 2x + 1}{(x - 1)(x^2 + 1)} ( x − 1 ) ( x 2 + 1 ) 3 x 2 − 2 x + 1 as a sum of simpler fractions:
A x − 1 + B x + C x 2 + 1 \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} x − 1 A + x 2 + 1 B x + C Thus, we set up the equation:
3 x 2 − 2 x + 1 ( x − 1 ) ( x 2 + 1 ) = A x − 1 + B x + C x 2 + 1 \frac{3x^2 - 2x + 1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1} ( x − 1 ) ( x 2 + 1 ) 3 x 2 − 2 x + 1 = x − 1 A + x 2 + 1 B x + C Step 2: Multiply both sides by the denominator
Multiply both sides of the equation by ( x − 1 ) ( x 2 + 1 ) (x - 1)(x^2 + 1) ( x − 1 ) ( x 2 + 1 )
3 x 2 − 2 x + 1 = A ( x 2 + 1 ) + ( B x + C ) ( x − 1 ) 3x^2 - 2x + 1 = A(x^2 + 1) + (Bx + C)(x - 1) 3 x 2 − 2 x + 1 = A ( x 2 + 1 ) + ( B x + C ) ( x − 1 ) Step 3: Solve for the constants A A A B B B C C C
Substitute x = 1 x = 1 x = 1 :
3 ( 1 ) 2 − 2 ( 1 ) + 1 = A ( 1 2 + 1 ) + ( B ( 1 ) + C ) ( 1 − 1 ) 3(1)^2 - 2(1) + 1 = A(1^2 + 1) + (B(1) + C)(1 - 1) 3 ( 1 ) 2 − 2 ( 1 ) + 1 = A ( 1 2 + 1 ) + ( B ( 1 ) + C ) ( 1 − 1 ) This simplifies to:
3 − 2 + 1 = A ( 2 ) ⇒ 2 A = 2 ⇒ A = 1 3 - 2 + 1 = A(2) \quad \Rightarrow \quad 2A = 2 \quad \Rightarrow \quad A = 1 3 − 2 + 1 = A ( 2 ) ⇒ 2 A = 2 ⇒ A = 1
Compare coefficients of powers of x x x :
Thus, we have:
A = 1 , B = 2 , C = 0 A = 1, \quad B = 2, \quad C = 0 A = 1 , B = 2 , C = 0 Step 4: Rewrite the integral
Now we can rewrite the integrand:
3 x 2 − 2 x + 1 ( x − 1 ) ( x 2 + 1 ) = 1 x − 1 + 2 x x 2 + 1 \frac{3x^2 - 2x + 1}{(x - 1)(x^2 + 1)} = \frac{1}{x - 1} + \frac{2x}{x^2 + 1} ( x − 1 ) ( x 2 + 1 ) 3 x 2 − 2 x + 1 = x − 1 1 + x 2 + 1 2 x Thus, the integral becomes:
∫ 2 3 1 x − 1 , d x + ∫ 2 3 2 x x 2 + 1 , d x \int_2^3 \frac{1}{x - 1} , dx + \int_2^3 \frac{2x}{x^2 + 1} , dx ∫ 2 3 x − 1 1 , d x + ∫ 2 3 x 2 + 1 2 x , d x Step 5: Integrate each term
First integral :
∫ 2 3 1 x − 1 , d x = ln ∣ x − 1 ∣ ∣ 2 3 = ln ( 3 − 1 ) − ln ( 2 − 1 ) = ln 2 − ln 1 = ln 2 \int_2^3 \frac{1}{x - 1} , dx = \ln|x - 1| \Big|_2^3 = \ln(3 - 1) - \ln(2 - 1) = \ln 2 - \ln 1 = \ln 2 ∫ 2 3 x − 1 1 , d x = ln ∣ x − 1∣ 2 3 = ln ( 3 − 1 ) − ln ( 2 − 1 ) = ln 2 − ln 1 = ln 2
Second integral :
∫ 2 3 2 x x 2 + 1 , d x = ln ( x 2 + 1 ) ∣ 2 3 = ln ( 3 2 + 1 ) − ln ( 2 2 + 1 ) = ln 10 − ln 5 \int_2^3 \frac{2x}{x^2 + 1} , dx = \ln(x^2 + 1) \Big|_2^3 = \ln(3^2 + 1) - \ln(2^2 + 1) = \ln 10 - \ln 5 ∫ 2 3 x 2 + 1 2 x , d x = ln ( x 2 + 1 ) 2 3 = ln ( 3 2 + 1 ) − ln ( 2 2 + 1 ) = ln 10 − ln 5 Step 6: Combine the results
Now combine the results of both integrals:
ln 2 + ln 10 − ln 5 \ln 2 + \ln 10 - \ln 5 ln 2 + ln 10 − ln 5 We can simplify this expression using the logarithmic property ln a + ln b = ln ( a b ) \ln a + \ln b = \ln(ab) ln a + ln b = ln ( ab )
ln ( 2 5 ) + ln 10 = ln ( 2 5 × 10 ) = ln 4 \ln\left(\frac{2}{5}\right) + \ln 10 = \ln\left(\frac{2}{5} \times 10\right) = \ln 4 ln ( 5 2 ) + ln 10 = ln ( 5 2 × 10 ) = ln 4 Thus, the value of the integral is:
ln 4 \boxed{\ln 4} ln 4
Partial Fraction Decomposition : Used to break down a complicated rational expression into simpler terms.Logarithmic Integration :
∫ 1 x − a , d x = ln ∣ x − a ∣ \int \frac{1}{x - a} , dx = \ln|x - a| ∫ x − a 1 , d x = ln ∣ x − a ∣ ∫ 2 x x 2 + 1 , d x = ln ( x 2 + 1 ) \int \frac{2x}{x^2 + 1} , dx = \ln(x^2 + 1) ∫ x 2 + 1 2 x , d x = ln ( x 2 + 1 )
Summary of Steps
Set up the partial fraction decomposition for the integrand.Multiply both sides by ( x − 1 ) ( x 2 + 1 ) (x - 1)(x^2 + 1) ( x − 1 ) ( x 2 + 1 ) Solve for constants A A A B B B C C C x = 1 x = 1 x = 1 Rewrite the integrand as the sum of simpler fractions.Integrate each term separately:
∫ 1 x − 1 , d x \int \frac{1}{x - 1} , dx ∫ x − 1 1 , d x ∫ 2 x x 2 + 1 , d x \int \frac{2x}{x^2 + 1} , dx ∫ x 2 + 1 2 x , d x
Combine the results to get the final answer: ln 4 \ln 4 ln 4
The final answer is:
ln 4 \boxed{\ln 4} ln 4 
