Question

Solve: \(\int \frac{3x - 5}{x^3 - x^2 - x + 1} dx\)

Hint:

When setting up partial fractions, always remember the rule for repeated factors: for a factor \((x-a)^k\), you need k terms: \(\frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + .... + \frac{A_k}{(x-a)^k}\). This is a common point of error.

Answer 1

Step 1: Understanding the Concept:
This is an indefinite integral of a rational function. The standard method for solving such integrals is to use partial fraction decomposition, which involves breaking down the complex fraction into a sum of simpler fractions that are easier to integrate.
Step 2: Key Formula or Approach:
1. Factor the denominator completely.
2. Express the rational function as a sum of partial fractions. For a denominator of the form \((x-a)^2(x-b)\), the decomposition is \(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}\).
3. Solve for the unknown coefficients (A, B, C).
4. Integrate the resulting simpler fractions using standard integration formulas: \(\int \frac{1}{u} du = \ln|u|\) and \(\int u^n du = \frac{u^{n+1}}{n+1}\).
Step 3: Detailed Explanation or Calculation:
1. Factor the denominator: \[ x^3 - x^2 - x + 1 \] Factor by grouping: \[ x^2(x - 1) - 1(x - 1) = (x^2 - 1)(x - 1) = (x - 1)(x + 1)(x - 1) = (x - 1)^2(x + 1) \] So the integral is: \[ \int \frac{3x - 5}{(x - 1)^2(x + 1)} dx \] 2. Decompose into partial fractions: \[ \frac{3x - 5}{(x - 1)^2(x + 1)} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{x + 1} \] Multiply both sides by the denominator \((x - 1)^2(x + 1)\): \[ 3x - 5 = A(x - 1)(x + 1) + B(x + 1) + C(x - 1)^2 \] 3. Solve for A, B, and C:
We can substitute convenient values of \(x\) to find the coefficients.
Let \(x = 1\): \[ 3(1) - 5 = A(0) + B(1 + 1) + C(0) \implies -2 = 2B \implies B = -1 \] Let \(x = -1\): \[ 3(-1) - 5 = A(0) + B(0) + C(-1 - 1)^2 \implies -8 = C(-2)^2 \implies -8 = 4C \implies C = -2 \] To find A, let's use another value, for example \(x = 0\): \[ 3(0) - 5 = A(0 - 1)(0 + 1) + B(0 + 1) + C(0 - 1)^2 \] \[ -5 = A(-1)(1) + B(1) + C(1) \implies -5 = -A + B + C \] Substitute the values of B and C we found: \[ -5 = -A + (-1) + (-2) \implies -5 = -A - 3 \implies -2 = -A \implies A = 2 \] 4. Integrate:
Now, substitute the coefficients back into the integral: \[ \int \left( \frac{2}{x - 1} - \frac{1}{(x - 1)^2} - \frac{2}{x + 1} \right) dx \] Integrate term by term: \[ = 2\int \frac{1}{x - 1} dx - \int \frac{1}{(x - 1)^2} dx - 2\int \frac{1}{x + 1} dx \] \[ = 2\ln|x - 1| - \int (x - 1)^{-2} dx - 2\ln|x + 1| \] \[ = 2\ln|x - 1| - \frac{(x - 1)^{-1}}{-1} - 2\ln|x + 1| + C \] \[ = 2\ln|x - 1| + \frac{1}{x - 1} - 2\ln|x + 1| + C \] Step 4: Final Answer:
The solution to the integral is: \[ 2\ln|x - 1| - 2\ln|x + 1| + \frac{1}{x - 1} + C \] This can also be written using logarithm properties as: \[ 2\ln\left|\frac{x - 1}{x + 1}\right| + \frac{1}{x - 1} + C \]