Question Statement

Evaluate the integral:

$\int \frac{5x^2-5x+2}{(x+1)(x-1{)}^2}dx$

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Background and Explanation

This problem involves integrating a rational function where the denominator contains both distinct and repeated linear factors. The method of partial fraction decomposition is used to break the complex fraction into simpler terms that can be integrated individually using standard logarithmic and power rules.

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Solution

Let $I=\int \frac{5x^2-5x+2}{(x+1)(x-1{)}^2}dx$

We begin by decomposing the rational function into partial fractions. Since the denominator has a distinct factor $(x+1)$ and a repeated factor $(x-1{)}^2$, we use the form:

$\frac{5x^2-5x+2}{(x+1)(x-1{)}^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1{)}^2}$

Multiplying both sides by $(x+1)(x-1{)}^2$ to clear the denominators:

$5x^2-5x+2=A(x-1{)}^2+B(x+1)(x-1)+C(x+1)$

Finding constant $A$:

Put $x=-1$ in equation (2):

$\begin{array}{rl}5(-1{)}^2-5(-1)+2& =A(-1-1{)}^2+B(-1+1)(-1-1)+C(-1+1)\\ 5(1)+5+2& =A(-2{)}^2+B(0)(-2)+C(0)\\ 5+5+2& =4A+0+0\\ 12& =4A\\ 3& =A\end{array}$

Finding constant $C$:

Put $x=1$ in equation (2):

$\begin{array}{rl}5(1{)}^2-5(1)+2& =A(1-1{)}^2+B(1+1)(1-1)+C(1+1)\\ 5-5+2& =A(0)+B(2)(0)+C(2)\\ 2& =0+0+2C\\ 1& =C\end{array}$

Finding constant $B$:

From equation (2), we expand and collect like terms:

$\begin{array}{rl}5x^2-5x+2& =A\left(x^2-2x+1\right)+B\left(x^2-1\right)+C(x+1)\\ 5x^2-5x+2& =A{x}^2-2Ax+A+B{x}^2-B+Cx+C\\ 5x^2-5x+2& =(A+B)x^2+(-2A+C)x+(A-B+C)\end{array}$

Comparing the coefficient of $x^2$ on both sides:

$\begin{array}{rl}5& =A+B\\ 5& =3+B\\ 2& =B\end{array}$

Rewriting the integrand:

Substituting $A=3$, $B=2$, and $C=1$ into equation (1):

$\frac{5x^2-5x+2}{(x+1)(x-1{)}^2}=\frac{3}{x+1}+\frac{2}{x-1}+\frac{1}{(x-1{)}^2}$

Integrating term by term:

$\begin{array}{rl}\int \frac{5x^2-5x+2}{(x+1)(x-1{)}^2}dx& =3\int \frac{1}{x+1}dx+2\int \frac{1}{x-1}dx+\int \frac{1}{(x-1{)}^2}dx\\ & =3\ln (x+1)+2\ln (x-1)+\int (x-1{)}^{-2}dx\\ & =3\ln (x+1)+2\ln (x-1)+\frac{(x-1{)}^{-2+1}}{-2+1}+c\\ & =3\ln (x+1)+2\ln (x-1)+\frac{(x-1{)}^{-1}}{-1}+c\\ & =3\ln (x+1)+2\ln (x-1)-\frac{1}{x-1}+c\end{array}$

Therefore, the final answer is:

$3\ln (x+1)+2\ln (x-1)-\frac{1}{x-1}+c$

(Note: In practice, absolute value signs are often included as $\ln |x+1|$ and $\ln |x-1|$ to ensure the logarithms are defined for all valid $x$ in the domain.)

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Key Formulas or Methods Used

• Partial Fraction Decomposition for distinct and repeated linear factors: $\frac{P(x)}{(x-a)(x-b{)}^2}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-b{)}^2}$

• Substitution method (Heaviside cover-up method) for finding constants by plugging in roots of the denominator factors

• Coefficient comparison method for finding remaining constants by equating coefficients of corresponding powers of $x$

• Standard integration formulas:

  • $\int \frac{1}{x}dx=\ln |x|+c$
  • $\int (x-a{)}^{n}dx=\frac{(x-a{)}^{n+1}}{n+1}+c$ for $n\ne -1$

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Summary of Steps

1. Set up the partial fraction decomposition with appropriate terms for the distinct factor $(x+1)$ and repeated factor $(x-1{)}^2$
2. Clear denominators by multiplying both sides by $(x+1)(x-1{)}^2$
3. Find $A$ by substituting $x=-1$ (the root of $x+1$) into the cleared equation
4. Find $C$ by substituting $x=1$ (the root of $x-1$) into the cleared equation
5. Find $B$ by expanding the right side and comparing coefficients of $x^2$ (or substitute another convenient $x$-value)
6. Rewrite the original integrand using the found constants $A=3$, $B=2$, $C=1$
7. Integrate each term separately: two logarithmic terms and one power rule term
8. Combine results and add the constant of integration $c$