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# Calculus

Integration of Functions

# Integration of Rational Functions

Page 1
Problem 1
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Problems 2-9

A rational function $$\large\frac{{P\left( x \right)}}{{Q\left( x \right)}}\normalsize,$$ where $${P\left( x \right)}$$ and $${Q\left( x \right)}$$ are both polynomials, can be integrated in four steps:

1. Reduce the fraction if it is improper (i.e. degree of $${P\left( x \right)}$$ is greater than degree of $${Q\left( x \right)}$$;
2. Factor $${Q\left( x \right)}$$ into linear and/or quadratic (irreducible) factors;
3. Decompose the fraction into a sum of partial fractions;
4. Calculate integrals of each partial fraction.

Consider these steps in more details.

### Step 1. Reducing an Improper Fraction

If the fraction is improper (i.e. degree of $${P\left( x \right)}$$ is greater than degree of $${Q\left( x \right)}$$), divide the numerator $${P\left( x \right)}$$ by the denominator $${Q\left( x \right)}$$ to obtain
${\frac{{P\left( x \right)}}{{Q\left( x \right)}} }={ F\left( x \right) + \frac{{R\left( x \right)}}{{Q\left( x \right)}},}$ where $$\large\frac{{R\left( x \right)}}{{Q\left( x \right)}}\normalsize$$ is a proper fraction.

### Step 2. Factoring $${Q\left( x \right)}$$ into Linear and/or Quadratic Factors

Write the denominator $${Q\left( x \right)}$$ as
${Q\left( x \right) } = {{\left( {x – a} \right)^\alpha } \cdots}\kern0pt{ {\left( {x – b} \right)^\beta }{\left( {{x^2} + px + q} \right)^\mu } \cdots}\kern0pt{ {\left( {{x^2} + rx + s} \right)^\nu },}$ where quadratic functions are irreducible, i.e. do not have real roots.

### Step 3. Decomposing the Rational Fraction into a Sum of Partial Fractions.

Write the function as follows:
${{\frac{{R\left( x \right)}}{{Q\left( x \right)}} }={ \frac{A}{{{{\left( {x – a} \right)}^\alpha }}} }\kern0pt{+ \frac{{{A_1}}}{{{{\left( {x – a} \right)}^{\alpha – 1}}}} + \ldots }}\kern0pt + {\frac{{{A_{\alpha – 1}}}}{{x – a}} + \ldots }\kern0pt + {\frac{B}{{{{\left( {x – b} \right)}^\beta }}} }+{ \frac{{{B_1}}}{{{{\left( {x – b} \right)}^{\beta – 1}}}} + \ldots }\kern0pt + {\frac{{{B_{\beta – 1}}}}{{x – b}} }\kern0pt + {\frac{{Kx + L}}{{{{\left( {{x^2} + px + q} \right)}^\mu }}} }+{ \frac{{{K_1}x + {L_1}}}{{{{\left( {{x^2} + px + q} \right)}^{\mu – 1}}}} + \ldots }\kern0pt + {\frac{{{K_{\mu – 1}}x + {L_{\mu – 1}}}}{{{x^2} + px + q}} + \ldots }\kern0pt + {\frac{{Mx + N}}{{{{\left( {{x^2} + rx + s} \right)}^\nu }}} }+{ \frac{{{M_1}x + {N_1}}}{{{{\left( {{x^2} + rx + s} \right)}^{\nu – 1}}}} + \ldots }\kern0pt + {\frac{{{M_{\nu – 1}}x + {N_{\nu – 1}}}}{{{x^2} + rx + s}}.}$ The total number of undetermined coefficients $${A_i},$$ $${B_i},$$ $${K_i},$$ $${L_i},$$ $${M_i},$$ $${N_i}, \ldots$$ must be equal to the degree of the denominator $${Q\left( x \right)}.$$

Then equate the coefficients of equal powers of $$x$$ by multiplying both sides of the latter expression by $${Q\left( x \right)}$$ and write the system of linear equations in $${A_i},$$ $${B_i},$$ $${K_i},$$ $${L_i},$$ $${M_i},$$ $${N_i}, \ldots$$ The resulting system must always have a unique solution.

### Step 4. Integrating partial fractions.

Use the following $$6$$ formulas to evaluate integrals of partial fractions with linear and quadratic denominators:
${1.\;\;}{\int {\frac{A}{{x – a}}dx} }={ \ln \left| {x – a} \right|}$ ${2.\;\;}{\int {\frac{A}{{{{\left( {x – a} \right)}^k}}}dx} }={ \frac{1}{{\left( {1 – k} \right){{\left( {x – a} \right)}^{k – 1}}}} }$ For fractions with quadratic denominators, first complete the square:
${\int {\frac{{Ax + B}}{{{{\left( {{x^2} + px + q} \right)}^k}}}dx} } = {\int {\frac{{At + B’}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}dt} ,}$ where $$t = x + {\large\frac{p}{2}\normalsize},$$ $${m^2} = {\large\frac{{4q – {p^2}}}{4}\normalsize},$$ $$B’ = B – {\large\frac{{Ap}}{2}\normalsize}.$$ Then use the formulas:
${3.\;\;}{\int {\frac{{tdt}}{{{t^2} + {m^2}}}} }={ \frac{1}{2}\ln \left( {{t^2} + {m^2}} \right)}$ ${4.\;\;}{ \int {\frac{{tdt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{1}{{2\left( {1 – k} \right){{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}} }$ ${5.\;\;}{\int {\frac{{dt}}{{{t^2} + {m^2}}}} }={ \frac{1}{a}\arctan \frac{t}{m}}$ The integral $$\large\int\normalsize {\large\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}\normalsize}$$ can be calculated in $$k$$ steps using the reduction formula:
${6.\;\;}{ \int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^k}}}} } = {\frac{t}{{2{m^2}\left( {k – 1} \right){{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}} } + {\frac{{2k – 3}}{{2{m^2}\left( {k – 1} \right)}} }\kern0pt{ \int {\frac{{dt}}{{{{\left( {{t^2} + {m^2}} \right)}^{k – 1}}}}} }$

## Solved Problems

Click on problem description to see solution.

### ✓Example 1

Evaluate the integral $${\large\int\normalsize} {{\large\frac{{2x + 3}}{{{x^2} – 9}}\normalsize} dx} .$$

### ✓Example 2

Evaluate $${\large\int\normalsize} {{\large\frac{{{x^2} – 2}}{{x + 1}}\normalsize} dx}.$$

### ✓Example 3

Evaluate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{{x^2} + 4x + 8}}\normalsize}.$$

### ✓Example 4

Evaluate the integral $${\large\int\normalsize} {\large\frac{{{x^2}dx}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)}}\normalsize} .$$

### ✓Example 5

Evaluate $${\large\int\normalsize} {\large\frac{{dx}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}\normalsize} .$$

### ✓Example 6

Find the integral $${\large\int\normalsize} {\large\frac{{dx}}{{{x^3} + 1}}\normalsize} .$$

### ✓Example 7

Evaluate the integral $${\large\int\normalsize} {\large\frac{{dx}}{{{x^4} – 1}}\normalsize} .$$

### ✓Example 8

Calculate the integral $${\large\int\normalsize} {{\large\frac{{5x}}{{{{\left( {x – 1} \right)}^3}}}\normalsize} dx}.$$

### ✓Example 9

Find the integral $${\large\int\normalsize} {\large\frac{{dx}}{{{{\left( {{x^2} + x – 1} \right)}^2}}}\normalsize} .$$

### Example 1.

Evaluate the integral $${\large\int\normalsize} {{\large\frac{{2x + 3}}{{{x^2} – 9}}\normalsize} dx} .$$

#### Solution.

We decompose the integrand into partial functions:
${\frac{{2x + 3}}{{{x^2} – 9}} } = {\frac{{2x + 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} } = {\frac{A}{{x – 3}} + \frac{B}{{x + 3}}.}$ Equate coefficients:
${{A\left( {x + 3} \right) }+{ B\left( {x – 3} \right) }={ 2x + 3,}\;\;}\Rightarrow {{Ax + 3A + Bx – 3B }={ 2x + 3,}\;\;}\Rightarrow {{\left( {A + B} \right)x + 3A – 3B }={ 2x + 3.}}$ Hence,
${\left\{ \begin{array}{l} A + B = 2\\ 3A – 3B = 3 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} A = \frac{3}{2}\\ B = \frac{1}{2} \end{array} \right..}$ Then
${\frac{{2x + 3}}{{{x^2} – 9}} } = {\frac{{\frac{3}{2}}}{{x – 3}} + \frac{{\frac{1}{2}}}{{x + 3}}.}$ The integral is equal to
${\int {\frac{{2x + 3}}{{{x^2} – 9}}dx} } = {{\frac{3}{2}\int {\frac{{dx}}{{x – 3}}} }+{ \frac{1}{2}\int {\frac{{dx}}{{x + 3}}} }} = {{\frac{3}{2}\ln \left| {x – 3} \right| }+{ \frac{1}{2}\ln \left| {x + 3} \right| }+{ C }} = {{\frac{1}{2}\ln \left| {{{\left( {x – 3} \right)}^2}\left( {x + 3} \right)} \right| }+{ C.}}$

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Problem 1
Page 2
Problems 2-9