Partial Fraction Integrals

Problem 1. Given a real number $\alpha$ , evaluate the integrals

$\displaystyle \int \frac{1}{x-\alpha}\, \mathrm dx \quad \text{and} \quad \int \frac{1}{(x-\alpha)^2}\, \mathrm dx.$

(Click for Solution)

Solution. By the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\ln|x-\alpha| &= \frac 1{x-\alpha} \\ \frac{\mathrm d}{\mathrm dx} \left(\frac 1{x-\alpha}\right) &= \frac{\mathrm d}{\mathrm dx} \left( (x-\alpha)^{-1} \right) \\ &= (-1)(x-\alpha)^{-2} \\ &= -\frac 1{(x-\alpha)^2}.\end{aligned}$

By linearity,

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left(-\frac 1{x-\alpha}\right) = - \frac{\mathrm d}{\mathrm dx} \left(\frac 1{x-\alpha}\right) = \frac 1{(x-\alpha)^2}.$

Therefore,

$\begin{aligned} \int \frac{1}{x-\alpha}\, \mathrm dx &= \ln|x-\alpha| + C_1, \\\int \frac{1}{(x-\alpha)^2}\, \mathrm dx &= -\frac 1{x-\alpha} + C_2. \end{aligned}$

Problem 2. Given distinct real numbers $\alpha,\beta$ , determine constants $A, B$ such that

$\displaystyle \frac{ 1 }{(x-\alpha)(x-\beta)} = \frac A{x-\alpha} + \frac B{x-\beta}.$

Hence, evaluate $\displaystyle \int \frac{ 1 }{ (x-\alpha)(x-\beta) } \, \mathrm dx$ .

(Click for Solution)

Solution. Firstly, multiply both sides by $x - \alpha$ :

$\displaystyle \frac 1{x-\beta} = A + \frac{B}{x-\beta} \cdot (x-\alpha).$

Setting $x = \alpha$ ,

$\displaystyle \frac 1{\alpha-\beta} = A + \frac{B}{\alpha-\beta} \cdot 0.$

Therefore, $A = 1/(\alpha - \beta)$ .

Similarly, multiply both sides by $x - \beta$ :

$\displaystyle \frac 1{x-\alpha} = \frac{ A }{ x - \alpha } \cdot (x-\beta) + B.$

Setting $x = \beta$ ,

$\displaystyle \frac 1{\beta-\alpha} = \frac{ A }{ \beta - \alpha } \cdot 0 + B.$

Therefore, $B = 1/(\beta-\alpha) = -A$ .

Consolidating,

$\begin{aligned} \frac{ 1 }{(x-\alpha)(x-\beta)} &= \frac {\frac 1{\alpha - \beta}}{x-\alpha} - \frac {\frac 1{\alpha - \beta}}{x-\beta} \\ &= \frac 1{\alpha - \beta}\left( \frac {1}{x-\alpha} - \frac {1}{x-\beta} \right). \end{aligned}$

Problem 3. Use Problems 1 and 2 to evaluate the integral

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} \, \mathrm dx.$

(Click for Solution)

Solution. Write

$\displaystyle \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} = \frac{x^2 + x+ 1}{x-2} \cdot \frac 1{(x-3)^2}.$

Write $x = (x-2)+2$ and expand the fraction therein:

$\begin{aligned} \frac{ x^2 + x + 1 }{x-2} &= \frac{ ((x-2)+2)^2 + ((x-2)+2) + 1}{x-2} \\ &= \frac{(x-2)^2 + 2 \cdot (x-2) \cdot 2 + 2^2 + (x-2) + 2 + 1}{x-2} \\ &= \frac{ (x-2)^2 + 5(x-2) + 7 }{x-2} \\ &= x-2 + 5 + \frac 7{x-2} \\ &= x+3 + \frac{7}{x-2} \\ &= (x-3)+6 + \frac{7}{x-2}. \end{aligned}$

Therefore, by multiplying excess terms,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} &= \left( (x-3)+6 + \frac{7}{x-2} \right) \cdot \frac 1{(x-3)^2} \\ &= \frac 1{x-3} + \frac{6}{(x-3)^2} + \frac 7{(x-2)(x-3)^2}. \end{aligned}$

By Question 2,

$\begin{aligned} \frac 1{(x-2)(x-3)} &= \frac 1{2-3} \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac 1{x-3} - \frac 1{x-2}. \end{aligned}$

Therefore,

$\begin{aligned} \frac 1{(x-2)(x-3)^2} &= \frac 1{x-3} \cdot \frac 1{(x-2)(x-3)} \\ &= \frac 1{x-3} \cdot \left(\frac 1{x-3} - \frac 1{x-2}\right) \\ &= \frac 1{(x-3)^2} - \frac 1{(x-2)(x-3)} \\ &= \frac 1{(x-3)^2} - \left(\frac 1{x-3} - \frac 1{x-2}\right) \\ &= \frac 1{x-2} - \frac 1{x-3} + \frac 1{(x-3)^2}. \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3)^2} &= \frac 1{x-3} + \frac{6}{(x-3)^2} + \frac 7{(x-2)(x-3)^2} \\ &= \frac 1{x-3} + \frac{6}{(x-3)^2} + 7 \cdot \left( \frac 1{x-2} - \frac 1{x-3} + \frac 1{(x-3)^2} \right) \\ &= \frac{7}{x-2} - \frac 6{x-3} + \frac{13}{(x-3)^2}.\end{aligned}$

By the linearity of integration and Question 1,

$\begin{aligned} &\int \frac{ x^2 + x + 1 }{(x-2)(x-3)^2}\, \mathrm dx \\ &= 7 \cdot \int \frac 1{x-2}\, \mathrm dx - 6 \cdot \int \frac 1{x-3}\, \mathrm dx + 13 \cdot \int \frac 1{(x-3)^2}\, \mathrm dx \\ &= 7 \ln|x-2| - 6 \ln |x-3| - \frac{13}{x-3} + C. \end{aligned}$

Remark 1. In the process of evaluating the integral, we have uncovered the partial fraction decomposition for the following expression with distinct $\alpha,\beta$ :

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x-\beta)^2} = \frac A{x-\alpha} + \frac {B}{x-\beta} + \frac C{(x-\beta)^2}.$

We leave it as an exercise to check that

$\begin{aligned} A &= \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)^2}, \\ B &= p -A = p - \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)^2}, \\ C &= \frac{p\beta^2 + q\beta + r }{\beta - \alpha}. \end{aligned}$

Problem 4. Similarly, evaluate

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} \, \mathrm dx.$

(Click for Solution)

Solution. Using the simplification in Problem 2,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} &= \frac{ x^2 + x + 1 }{(x-3) (x-4)} \cdot \frac 1{(x-3)(x-4)} \\ &= \left((x-3)+6 + \frac{7}{x-2}\right) \cdot \frac 1{(x-3)(x-4)} \\ &= \frac{1}{x-4} + \frac 6{(x-3)(x-4)} + \frac 7{(x-2)(x-3)(x-4)}. \end{aligned}$

By Question 2 applied twice,

$\begin{aligned} \frac 1{(x-3)(x-4)} &= \frac 1{3-4} \left( \frac 1{x-3} - \frac 1{x-4} \right) \\ &= \frac 1{x-4} - \frac{1}{x-3} \end{aligned}$

so that

$\begin{aligned} \frac 1{(x-2)(x-3)(x-4)} &= \frac 1{x-2} \cdot \frac 1{(x-3)(x-4)} \\ &= \frac 1{x-2} \left( \frac 1{x-4} - \frac{1}{x-3} \right) \\ &= \frac 1{(x-2)(x-4)} - \frac{1}{(x-2)(x-3)} \\ &= \frac 1{2-4} \left(\frac 1{x-2} - \frac 1{x-4}\right) - \frac 1{2-3} \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac 12\left(\frac 1{x-4} - \frac 1{x-2}\right) + \left(\frac 1{x-2} - \frac 1{x-3}\right) \\ &= \frac{\frac 12}{x-2} - \frac 1{x-3} + \frac {\frac 12}{x-4}. \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)} &= \frac{ x^2 + x + 1 }{(x-3) (x-4)} \cdot \frac 1{(x-3)(x-4)} \\ &= \frac{\frac 72}{x-2} - \frac {13}{x-3}+ \frac{\frac{21}2}{x-4} .\end{aligned}$

By the linearity of integration and Question 1,

$\begin{aligned} \int \frac{ x^2 + x + 1 }{(x-2)(x-3) (x-4)}\, \mathrm dx &= \frac 72 \ln|x-2| - 13 \ln|x-3| + \frac{21}2 \ln|x-4| + C.\end{aligned}$

Remark 2. The corresponding partial fraction decomposition with distinct $\alpha,\beta,\gamma$ is as follows:

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x-\beta)(x-\gamma)} = \frac A{x-\alpha} + \frac {B}{x-\beta} + \frac C{x-\gamma}.$

We leave it as an exercise to check that

$\begin{aligned} A &= \frac{ p\alpha^2 + q\alpha + r }{(\alpha - \beta)(\alpha - \gamma)}, \\ B &= \frac{ p\beta^2 + q\beta + r }{(\beta - \alpha)(\beta - \gamma)}, \\ C &= \frac{ p\gamma^2 + q\gamma + r }{(\gamma - \alpha)(\gamma - \beta)}. \end{aligned}$

Problem 5. Given distinct real numbers $\alpha,\beta$ , determine constants $A,B,C$ such that

$\displaystyle \frac{ \alpha^2+\beta^2 }{(x-\alpha)(x^2+\beta^2) } = \frac A{x-\alpha} + \frac {Bx+C}{x^2 + \beta^2}.$

Hence, evaluate

$\displaystyle \int \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } \, \mathrm dx.$

(Click for Solution)

Solution. By observing that $0 = x^2 -x^2$ ,

$\begin{aligned} \frac{\alpha^2 + \beta^2}{(x-\alpha)(x^2 +\beta^2)} &= \frac{x^2 + \beta^2 - (x^2 - \alpha^2) }{(x-\alpha)(x^2+\beta^2)} \\ &= \frac{x^2 + \beta^2}{(x-\alpha)(x^2+\beta^2)} - \frac{x^2 -\alpha^2}{(x-\alpha)(x^2+\beta^2)} \\ &= \frac 1{x-\alpha} - \frac{(x-\alpha)(x+\alpha)}{(x-\alpha)(x^2+\beta^2)} \\ &= \frac 1{x-\alpha} - \frac{x+\alpha}{x^2+\beta^2}, \end{aligned}$

so that $A = 1, B = -1, C = -\alpha$ . Dividing both sides by $\alpha^2 +\beta^2$ for the next part,

$\displaystyle \frac 1{(x-\alpha)(x^2+\beta^2)} = \frac 1{\alpha^2+\beta^2} \left( \frac 1{x-\alpha} - \frac{x+\alpha}{x^2+\beta^2} \right).$

Using the solution in Problem 3,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } &= \frac{ x^2 + x + 1 }{x-2 } \cdot \frac 1{x^2+9} \\ &= \left(x+3 + \frac 7{x-2}\right)\cdot \frac 1{x^2+9} \\ &= \frac{x+3}{x^2+9} + \frac 7{(x-2)(x^2+9)}. \end{aligned}$

Using the first part with $\alpha = 2$ and $\beta = 3$ ,

$\begin{aligned} \frac 1{(x-2)(x^2+9)} &= \frac 1{2^2 + 3^2} \left(\frac 1{x-2} - \frac{x + 2}{x^2 + 3^2}\right) \\ &= \frac 1{13} \left(\frac 1{x-2} - \frac{x+2}{x^2+9}\right). \end{aligned}$

Combining the displays,

$\begin{aligned} \frac{ x^2 + x + 1 }{(x-2)(x^2+9) } &= \frac{x+3}{x^2+9} + \frac 7{(x-2)(x^2+9)} \\ &= \frac{x+3}{x^2+9} + \frac 7{13} \left(\frac 1{x-2} - \frac{x+2}{x^2+9}\right) \\ &= \frac 1{13} \left( \frac{7}{x-2} + \frac{13(x+3) - 7(x+2)}{x^2+9} \right) \\ &= \frac 1{13} \left( \frac{7}{x-2} + \frac{ 6x+ 25}{x^2+9} \right). \end{aligned}$

Using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} \ln(x^2 + 9) &= \frac{2x}{x^2+9}, \\ \frac{\mathrm d}{\mathrm dx} \tan^{-1}(x/3) &= \frac{1}{1 + (x/3)^2} \cdot \frac 13 = \frac{3}{x^2+9}. \end{aligned}$

Therefore,

$\begin{aligned} \int \frac{x}{x^2+9}\, \mathrm dx &= \frac 12 \ln(x^2+9) + C_1, \\ \int \frac{1}{x^2+9}\, \mathrm dx &= \frac 13 \tan^{-1}(x/3) + C_2, \end{aligned}$

By the linearity of integration,

$\begin{aligned} \int \frac{x^2+x+1}{(x-2)(x^2+9)}\, \mathrm dx &= \frac 1{13} \int \left( \frac{7}{x-2} + \frac{ 6x}{x^2+9}+ \frac{ 25}{x^2+9} \right)\, \mathrm dx \\ &= \frac 1{13}\left( \ln|x-2| + \frac 62 \ln(x^2 + 9) + \frac{25}{3} \tan^{-1}(x/3) \right) + C \\ &= \frac 1{13} \ln|x-2| + \frac{3}{13} \ln(x^2+9) + \frac{25}{39} \tan^{-1}(x/3) + C. \end{aligned}$

Remark 3. The corresponding partial fraction decomposition with distinct $\alpha,\beta$ is as follows:

$\displaystyle \frac{px^2 + qx + r}{(x-\alpha)(x^2+\beta^2)} = \frac A{x-\alpha} + \frac {Bx+C}{x^2 + \beta^2}.$

—Joel Kindiak, 29 Mar 26, 1546H

KindiakMath

Partial Fraction Integrals

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Partial Fraction Integrals

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