AOS4 Topic 7: Partial Fractions

Introduction

Partial fractions, a method of algebraic manipulation, operate as a link between the complicated and the understandable. Partial fractions allow mathematicians to confidently and precisely navigate the maze of calculus by breaking down complex rational functions into simpler components.


Definition

If \(g(x)\) and \(h(x)\) are polynomials, then \(f(x) = \frac{g(x)} {h(x)}\) is a rational function; e.g. \((x) = \frac{4x + 2} {x^2 − 1}\).

If the degree of \(g(x)\) is less than the degree of \(h(x)\), then \(f(x)\) is a proper fraction.

If the degree of \(g(x)\) is greater than or equal to the degree of \(h(x)\), then \(f(x)\) is an improper fraction.

A rational function may be expressed as a sum of simpler functions by resolving it into what are called partial fractions. For example:

\(\frac{4x + 2}{x^2 − 1} = \frac{3}{x − 1} + \frac{1}{x + 1}\)


Types of Partial Fractions

There are two types of fractions as follows:

1-Proper Partial Fractions

2-Improper Partial Fractions



1-Proper Partial Fractions

In the field of fractions that are partial, a fraction is deemed proper if the degree of the numerator is strictly less than the degree of the denominator.


Example 1

Consider the rational function:

\(f(x) = \frac{3x + 2}{x^2 + x + 1}\)

In this case, the degree of the numerator (which is 1) is strictly less than the degree of the denominator (which is 2). Therefore, the function \(f(x)\) is a proper fraction.


Example 2

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

In this instance, the degree of the numerator (which is 2) is strictly less than the degree of the denominator (which is 3). Hence, \(g(x)\) is also a proper fraction.



Rules for Proper Partial Fraction Decomposition


Rule 1: Linear Factor

For every linear factor \((ax + b)\) in the denominator, there will be a partial fraction of the form \(\frac{A}{ax + b}\)

Example:

Consider the rational function:

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

Here, the denominator contains a linear factor \((2x - 5)\). So, we decompose it into partial fractions:

\(f(x) = \frac{A}{2x - 5}\)

To find \(A\), we can multiply both sides of the equation by \((2x - 5)\) and then solve for \(A\).


Rule 2: Repeated Linear Factor

For every repeated linear factor \((ax + b)^2\) in the denominator, there will be partial fractions of the form \(\frac{A}{ax + b} + \frac{B}{(ax + b)^2}\).

Example:

Consider the rational function:

\(g(x) = \frac{2x + 3} {x^2 + 2x + 1}\)

Here, the denominator contains a repeated linear factor \((x + 1)^2\). So, we decompose it into partial fractions:

\(g(x) = \frac{A}{x + 1} + \frac{B}{(x + 1)^2}\)

To find \(A\) and \(B\), we can multiply both sides of the equation by \((x + 1)\) and \(((x + 1)^2)\) respectively, and then solve for \(A\) and \(B\).


Rule 3: Irreducible Quadratic Factor

For every irreducible quadratic factor \((ax^2 + bx + c)\) in the denominator, there will be a partial fraction of the form \(\frac{Ax + B}{ax^2 + bx + c}\).

Example:

Consider the rational function:

\(h(x) = \frac{3x + 2}{x^2 + x + 2}\)

Here, the denominator contains an irreducible quadratic factor \((x^2 + x + 2)\). So, we decompose it into partial fractions:

\(h(x) = \frac{Ax + B}{x^2 + x + 2}\)

To find \(A\) and \(B\), we can use various techniques such as comparing coefficients or substituting values.


Rule 4: Repeated Irreducible Quadratic Factor

For every repeated irreducible quadratic factor \((ax^2 + bx + c)^2\) in the denominator, there will be partial fractions of the form \(\frac{Ax + B}{ax^2 + bx + c} + \frac{Cx + D}{(ax^2 + bx + c)^2}\).

Example:

Consider the rational function:

\(k(x) = \frac{4x + 3}{x^2 + 2x + 1)^2}\)

Here, the denominator contains a repeated irreducible quadratic factor \((x^2 + 2x + 1)^2\). So, we decompose it into partial fractions:

\(k(x) = \frac{Ax + B}{x^2 + 2x + 1} + \frac{Cx + D}{(x^2 + 2x + 1)^2)}\)

To find \(A\), \(B\), \(C\), and \(D\), we can multiply both sides of the equation by \((x^2 + 2x + 1)\) and \((x^2 + 2x + 1)^2\) respectively, and then solve for \(A\), \(B\), \(C\), and \(D\).


Note1:

A quadratic expression is irreducible if it cannot be factorized over ℝ (the set of real numbers), that is, if its discriminant is negative. For example, both \((x^2 + 1)\) and \((x^2 + 4x + 10)\) are irreducible.


Note2:

If \(f(x) = \frac{g(x)}{h(x)}\) is an improper fraction, i.e., if the degree of \(g(x)\) is greater than or equal to the degree of \(h(x)\), then the division must be performed first.


Rules for Partial Fraction Decomposition

To resolve an algebraic fraction into its partial fractions:

Step 1:

Factorize the numerator and denominator and simplify the rational expression before doing partial fraction decomposition.

Step 2:

Split the rational expression as per the formula for partial fractions. \(\frac{P}{(ax + b)^2} = \frac{A}{ax + b} + \frac{B}{(ax + b)^2}\). There are different partial fraction formulas based on the numerator and denominator expression.

Step 3:

Take the LCM of the factors of the denominators of the partial fractions, and multiply both sides of the equation with this LCM.

Step 4:

Simplify and obtain the values of A and B by comparing coefficients of like terms on both sides.

Step 5:

Substitute the values of the constants A and B on the right side of the equation to obtain the partial fraction.




Improper Partial Fraction

In the realm of partial fractions, a fraction is considered improper if the degree of the numerator is greater than or equal to the degree of the denominator.

Example 1

Consider the rational function:

\(f(x) = \frac{2x^2 + 3x + 1}{x^2 + x + 1}\)

In this case, the degree of the numerator (which is 2) is greater than the degree of the denominator (which is 2). Therefore, the function \(f(x)\) is an improper fraction.

Example 2

\(g(x) = \frac{4x^3 + 2x^2 + 3}{2x^2 + 1}\)

In this instance, the degree of the numerator (which is 3) is greater than the degree of the denominator (which is 2). Hence, \(g(x)\) is also an improper fraction.


Using partial fractions for integration

Using partial fractions for integration involves breaking down a complex rational function into simpler fractions, termed partial fractions. This simplification enables the integration of the rational function by transforming it into a sum of integrals that are easier to compute.

The process of using partial fractions for integration typically involves the following steps:

Decomposition: The given rational function is decomposed into its constituent partial fractions. This step involves expressing the rational function as a sum of fractions, each with a simpler denominator, such as linear factors or irreducible quadratic factors.

Integration of Partial Fractions: Each partial fraction is integrated separately. The integrals of these individual fractions are typically straightforward to compute using basic integration techniques such as power rule integration, natural logarithm integration, or trigonometric substitution.

Combination of Integrals: After integrating each partial fraction, the resulting integrals are combined to obtain the final integral of the original rational function.


Strategies for Partial Fraction Integration

Distinct linear factors

\(\int \frac{3x + 5}{(x - 1)(x + 3)} dx\)

Improper fractions

If the degree of the numerator is greater than or equal to the degree of the denominator, then division must take place first.

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

Repeated linear factor

\(\int \frac{3x + 1}{(x + 2)^2} dx\)

Irreducible quadratic factor

\(\int \frac{4}{(x + 1)(x^2 + 1)} dx\)



Practice using the simulation below.

Created with GeoGebra®, by Gabriel Cheow, Link

Example 1

Find the partial fraction decomposition of the rational function:

\(\frac{5x + 1}{(x - 1)(x + 2)}\)

Solution:

Solution:

To decompose the rational function into partial fractions, we start by writing:

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

Now, we clear the denominators by multiplying both sides by \((x - 1)(x + 2)\):

\((x - 1)(x + 2) \cdot \frac{5x + 1}{(x - 1)(x + 2)} = (x - 1)(x + 2) \cdot \left( \frac{A}{x - 1} + \frac{B}{x + 2} \right)\)

This simplifies to:

\(5x + 1 = A(x + 2) + B(x - 1)\)

To solve for \(A\) and \(B\), we can substitute suitable values for \(x\) that eliminate one of the terms. Let's choose \(x = 1\) and \(x = -2\):

At \(x = 1\): \(6 = 3A\), so \(A = 2\)

At \(x = -2\): \(-9 = -3B\), so \(B = 3\)

Therefore, the partial fraction decomposition of \(\frac{5x + 1}{(x - 1)(x + 2)}\) is:

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

This represents the rational function as a sum of simpler fractions.

Example 2

Resolve the following rational expression into partial fraction:

\(\frac{-1}{(x + 1)(2x + 1)}\)

Solution:

Given the rational function \(\frac{-1}{(x + 1)(2x + 1)}\), let's decompose it into partial fractions:

\(\frac{-1}{(x + 1)(2x + 1)} = \frac{A}{x + 1} + \frac{B}{2x + 1}\)

To find the values of \(A\) and \(B\), we'll clear the denominators by multiplying both sides by \((x + 1)(2x + 1)\):

\((x + 1)(2x + 1) \cdot \frac{-1}{(x + 1)(2x + 1)} = (x + 1)(2x + 1) \cdot \left( \frac{A}{x + 1} + \frac{B}{2x + 1} \right)\)

After simplification, we get:

\(-1 = A(2x + 1) + B(x + 1)\)

To solve for \(A\) and \(B\), we can substitute suitable values for \(x\) that eliminate one of the terms. Let's choose \(x = -1\) and \(x = \frac{-1}{2}\):

At \(x = -1\): \(-1 = A(2(-1)+1)+B(-1+1)\): \(-1 = -1(A)+B(0)\), So \(A=1\)

At \(x = \frac{-1}{2}\): \(-1 = A(2(\frac{-1}{2})+1)+B(\frac{-1}{2}+1)\): \(-1 = A(0)+B(\frac{1}{2})\), So \(B=-2\)

Therefore, the partial fraction decomposition of \(\frac{-1}{(x + 1)(2x + 1)}\) is:

\(\frac{-1}{(x + 1)(2x + 1)} = \frac{1}{x + 1} - \frac{2}{2x + 1}\)

This represents the rational function as a sum of simpler fractions.

Example 3

Resolve \(\frac{2x + 10}{(x + 1)(x - 1)^2}\) into partial fraction.

Solution:

Since the denominator has a repeated linear factor and a single linear factor, there are three partial fractions:

\(\frac{2x + 10}{(x + 1)(x - 1)^2} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{(x - 1)^2}\)

This gives the equation:

\(2x + 10 = A(x - 1)^2 + B(x + 1)(x - 1) + C(x + 1)\)


Let \(x = 1\): \(12 = 2C\)
∴ \(C = 6\)

Let \(x = -1\): \(8 = 4A\)
∴ \(A = 2\)

Let \(x = 0\): \(10 = A - B + C\)
∴ \(B = A + C - 10 = -2\)

Hence:

\(\frac{2x + 10}{(x + 1)(x - 1)^2} = \frac{2}{x + 1} - \frac{2}{x - 1} + \frac{6}{(x - 1)^2}\)


Example 4

Resolve the following rational expression into partial fractions:

\(\frac{2x - 2}{(x + 1)(x - 2)^2}\)


Solution:

Since the denominator has a repeated linear factor \((x - 2)^2\) and a single linear factor \((x + 1)\), we'll have three partial fractions:

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

Clearing Denominators:

Next, we clear the denominators by multiplying both sides of the equation by \((x + 1)(x - 2)^2\):

\((x + 1)(x - 2)^2 \cdot (2x - 2) = (x + 1)(x - 2)^2 \cdot \left( \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{(x - 2)^2} \right)\)

After simplification:

\(2x - 2 = A(x - 2)^2 + B(x + 1)(x - 2) + C(x + 1)\)


Now, we can solve for the constants \(A\), \(B\), and \(C\).

Substituting \(x = -1\):

\(2(-1) - 2 = A((-1) - 2)^2 + B((-1) + 1)((-1) - 2) + C((-1) + 1)\)

\(-4 = A(-3)^2 + 0 + 0\)

\(-4 = 9A\)
\(A = -\frac{4}{9}\)

Substituting \(x = 2\):

\(2(2) - 2 = A((2) - 2)^2 + B((2) + 1)((2) - 2) + C((2) + 1)\)

\(4 - 2 = A(0) + B(0) + C(3)\)

\(2 = 3C\)
\(C = \frac{2}{3}\)

Substituting \(x = 0\):

\(2(0) - 2 = -\frac{4}{9}(0 - 2)^2 + B(0 + 1)(0 - 2) + \frac{2}{3}(0 + 1)\)

\(-2 = -\frac{4}{9}(2)^2 + B(1)(-2) + \frac{2}{3}(1)\)

\(-2 = -\frac{16}{9} - 2B + \frac{2}{3}\)
\(B = \frac{4}{9}\)

Therefore, the correct values for \(A\), \(B\), and \(C\) are:

\(A = -\frac{4}{9}\)

\(B = \frac{4}{9}\)

\(C = \frac{2}{3}\)

Hence:

\(\frac{2x -2}{(x + 1)(x - 2)^2} = \frac{-4}{9(x + 1)} + \frac{4}{9(x - 2)} + \frac{2}{2(x - 2)^2}\)


Example 5

Resolve \( \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} \) into partial fractions.

Solution:

Since the denominator has a single linear factor and an irreducible quadratic factor (i.e., cannot be reduced to linear factors), there are two partial fractions:

\( \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + x + 1} \)

This gives the equation:

\( x^2 + 6x + 5 = A(x^2 + x + 1) + (Bx + C)(x - 2) \) (1)

Substituting \( x = 2 \):

\( 2^2 + 6(2) + 5 = A(2^2 + 2 + 1) \)
\( 21 = 7A \)
\( \therefore A = 3 \)

We can rewrite equation (1) as:

\( x^2 + 6x + 5 = A(x^2 + x + 1) + (Bx + C)(x - 2) \)
\( = A(x^2 + x + 1) + Bx^2 - 2Bx + Cx - 2C \)
\( = (A + B)x^2 + (A - 2B + C)x + (A - 2C) \)


Since \( A = 3 \), this gives:

\( x^2 + 6x + 5 = (3 + B)x^2 + (3 - 2B + C)x + (3 - 2C) \)

Equate coefficients:

\( 3 + B = 1 \) and \( 3 - 2C = 5 \)
\( \therefore B = -2 \) and \( \therefore C = -1 \)

Check: \( 3 - 2B + C = 3 - 2(-2) + (-1) = 6 \)

Therefore,

\( \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} = \frac{3}{x - 2} - \frac{2x + 1}{x^2 + x + 1} \)

Note: The values of \( B \) and \( C \) could also be found by substituting \( x = 0 \) and \( x = 1 \) in equation (1).

Example 6

Resolve \( \frac{x^2 + 2x - 13}{2x^3 + 6x^2 + 2x + 6} \) into partial fractions.


Solution:

Since the denominator is a cubic polynomial, we need to factorize it first.

Given denominator: \( 2x^3 + 6x^2 + 2x + 6 \)

Factorizing by grouping:

\( 2x^2(x + 3) + 2(x + 3) \)

Now, we can see that the common factor is \( x + 3 \), so we can rewrite the denominator as:

\( (x + 3)(2x^2 + 2) \)

Now, we have the denominator in factored form. We can now perform partial fraction decomposition.

We want to express the given expression as:

\( \frac{x^2 + 2x - 13}{2x^3 + 6x^2 + 2x + 6} = \frac{A}{x + 3} + \frac{Bx + C}{2x^2 + 2} \)

Now, let's find the values of \( A \), \( B \), and \( C \).

Clearing Denominators:

Multiplying both sides by \( (x + 3)(2x^2 + 2) \):

\( (x + 3)(2x^2 + 2) \cdot \frac{x^2 + 2x - 13}{(x+3)(2x^2+2)} = (x + 3)(2x^2 + 2) \cdot \left( \frac{A}{x + 1} + \frac{Bx + C}{2x^2 + 2} \right) \)

After simplification, we get:

\( x^2 + 2x - 13 = A(2x^2 + 2) + (Bx + C)(x + 3) \) (1)

Now, we can solve for \( A \), \( B \), and \( C \).

At \(x=-3: 9-6+3=A(18+2)+0 : 6=20A\)

So, \(A= \frac{3}{10}\)

After simplification equation 1, we get:

\( x^2 + 2x - 13 = A(2x^2 + 2) + (Bx + C)(x + 3) \)

Now, let's equate the coefficients of like terms:

For \(x^2\): \(1 = 2A + B\)

Since, \(A= \frac{3}{10}\) So, \(B= \frac{2}{5}\)

For \(x\): \(2 = 3B + C\)

Since, \(B= \frac{2}{5}\) So, \(C= \frac{11}{10}\)

Therefore, the partial fraction decomposition of the given expression is:

\( \frac{x^2 + 2x - 13}{2x^3 + 6x^2 + 2x + 6} = \frac{\frac{3}{10}}{x + 3} + \frac{\frac{2}{5}x + \frac{11}{10}}{2x^2 + 2} \)

\( \frac{x^2 + 2x - 13}{2x^3 + 6x^2 + 2x + 6} = \frac{3}{10(x + 3)} +\frac{4x+11}{10(2x^2 + 2)} \)

Example 7

Express \( \frac{x^5 + 2}{x^2 - 1} \) as partial fractions:

Solution

Dividing through:

$$ \frac{x^5 + 2}{x^2 - 1} = \frac{x^5 + 0x^4 + 0x^3 + 0x^2 + 0x + 2}{x^2 - 1} $$

By dividing we get:

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

By expressing \( \frac{x + 2}{x^2 - 1} = \frac{x+2}{(x-1)(x+1)} \) as partial fractions, we obtain:

\(A=\frac{-1}{2}\) and \(B=\frac{3}{2}\)

Therefore

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


Example 8

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

Solution:

From previous example we have:

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

Hence

$$ \int \frac{x^5 + 2}{x^2 - 1} \, dx = \int \left( x^3 +x -\frac{1}{2(x + 1)} + \frac{3}{2(x - 1)} \right) \, dx $$

$$ = \frac{x^4}{4} + \frac{x^2}{2} - \frac{1}{2} \ln{|x + 1|} + \frac{3}{2} \ln{|x - 1|} + c $$

$$ = \frac{x^4}{4} + \frac{x^2}{2} + \frac{1}{2} \ln{\left| \frac{(x - 1)^3}{(x + 1)} \right|} + c $$

Example 9

Express $$ \frac{3x + 1}{(x + 2)^2} $$ in partial fractions and hence find $$ \int \frac{3x + 1}{(x + 2)^2} \, dx $$

Solution

$$ \frac{3x + 1}{(x + 2)^2} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} $$

Write 2 Then \(3x + 1 = A(x + 2) + B\)

Substituting \(x = -2\) gives \( -5 = B \).

Substituting \(x = 0\) gives \( 1 = 2A + B \) and therefore \( A = 3 \).

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

Thus

$$ \int \frac{3x + 1}{(x + 2)^2} \, dx = \int \left( \frac{3}{x + 2} - \frac{5}{(x + 2)^2} \right) \, dx $$

$$ = 3 \ln{|x + 2|} + \frac{5}{x + 2} + c $$


Example 10

Find an antiderivative of \( \frac{4}{(x + 1)(x^2 + 1)} \) by first expressing it as partial fractions.

Solution

Write \( \frac{4}{(x + 1)(x^2 + 1)} = \frac{A}{x + 1} + \frac{Bx + C}{x^2 + 1} \)

Then \( 4 = A(x^2 + 1) + (Bx + C)(x + 1) \)

Let \( x = -1 \): \( 4 = 2A \) therefore \( A = 2 \)

Let \( x = 0 \): \( 4 = A + C \) therefore \( C = 2 \)

Let \( x = 1 \): \( 4 = 2A + 2(B + C) \) therefore \( B = -2 \)

Hence \( \frac{4}{(x + 1)(x^2 + 1)} = \frac{2}{x + 1} + \frac{2 - 2x}{x^2 + 1} \)

We now turn to the integration:

\( \int \frac{4}{(x + 1)(x^2 + 1)} \, dx = \int \frac{2}{x + 1} + \frac{2 - 2x}{x^2 + 1} \, dx \)

\( = \int \frac{2}{x + 1} \, dx + \int \frac{2}{x^2 + 1} \, dx - \int \frac{2x}{x^2 + 1} \, dx \)

\( = 2 \ln{|x + 1|} + 2 \arctan x - \ln{(x^2 + 1)} + c \)

\( = \ln{\left( \frac{(x + 1)^2}{x^2 + 1} \right)} + 2 \arctan x + c \)


Example 11

Decompose \( \frac{5}{(x - 4)(x^2 + 2)} \) into partial fractions and find its antiderivatives.

Solution:

First, we decompose the rational function into partial fractions:

$$ \frac{5}{(x - 4)(x^2 + 2)} = \frac{A}{x - 4} + \frac{Bx + C}{x^2 + 2} $$

By multiplying with L.C.M \((x-4)(x^2+2)\) we get:

\(5 = A(x^2 + 2) + Bx + C(x - 4) \)

At \(x=4\): \(A= \frac{5}{18}\)

By comparing the co-efficients of \(x^2\): we get

\( 0= A+B \) So, \(B=\frac{-5}{18} \)

By comparing the co-efficients of \(x\): we get

\( 0= -4B+C \) So, \(C=\frac{-10}{9} \)

Hence

$$ \frac{5}{(x - 4)(x^2 + 2)} = \frac{5}{18(x - 4)} - \frac{5(x-4)}{18(x^2 + 2)} $$

Now we turn to the integration:

\( \int \frac{5}{(x - 4)(x^2 + 2)} \, dx = \int \frac{5}{18(x - 4)} - \frac{5(x-4)}{18(x^2 + 2)} \, dx \)

\( = \int \frac{5}{18(x - 4)} \, dx - \int \frac{5(x-4)}{18(x^2+2)} \, dx \)

\( = \frac{5}{18} \int \frac{1}{x - 4} \, dx - \frac{5}{18} \int \frac{x-4}{(x^2+2)} \, dx \)

\( \int \frac{1}{x - 4} \, dx = \ln{|x - 4|}\)

\(\int \frac{x-4}{(x^2+2)} \, dx= \frac{\ln(x^2+2)}{2} - 2^\frac{3}{2} \arctan\left(\frac{x}{\sqrt{2}}\right) + C \)

Hence

\( \int \frac{5}{(x - 4)(x^2 + 2)} \, dx = \frac{5}{18} (\ln{|x - 4|}+ \frac{\ln(x^2+2)}{2} - 2^\frac{3}{2} \arctan\left(\frac{x}{\sqrt{2}}\right) + C) \)

Example 12

Let \( f(x) = \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} \).

  1. a) Express \( f(x) \) as partial fractions.
  2. b) Hence find an antiderivative of \( f(x) \).
  3. c) Hence evaluate \( \int_{-2}^{-1} f(x) \, dx \).

Solution:

  • Partial fractions for each factor:

    • \( \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} = \frac{A}{x - 2} + \frac{Bx + C}{x^2 + x + 1} \)

  • Multiply through by the common denominator of \( (x - 2)(x^2 + x + 1) \):

    • \( x^2 + 6x + 5 = A(x^2 + x + 1) + (Bx + C)(x - 2) \)

    \( x^2 + 6x + 5 = Ax^2 + Ax + A + Bx^2 - 2Bx + Cx - 2C \)

    At x=2: \(A=3\)

    By comparing the coefficients of \(x^2\) and \(x\):

    we have: \(B=-2\) and \(C=-1\)

    Hence

    \( \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} = \frac{3}{x - 2} - \frac{2x + 1}{x^2 + x + 1} \)

    \(\int \left(\frac{3}{x-2} - \frac{2x + 1}{x^2 + x + 1}\right)dx\)

    Apply linearity:

    \( = 3\int \frac{1}{x - 2} dx - \int \frac{2x + 1}{x^2 + x + 1} dx \)

    Now solving:

    \( \int \frac{1}{x - 2} dx \)

    Substitute \( u = x - 2 \) \( \rightarrow \, du = dx \) (steps):

    \( = \int \frac{1}{u} du \)

    This is a standard integral:

    \( = \ln(|u|) \)

    \( = \ln(|x - 2|) \)

    Now solving:

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

    Substitute \( u = x^2 + x + 1 \) \( \rightarrow \, du = (2x + 1) dx \) (steps):

    \( = \int \frac{1}{u} du \)

    Use previous result:

    \(= \ln(|u|)\)

    \(= \ln(|x^2 + x + 1|)\)

    Plug in solved integrals:

    \(\int \left(\frac{3}{x-2} - \frac{2x + 1}{x^2 + x + 1}\right)dx = 3\ln(|x - 2|) - \ln(|x^2 + x + 1|) + C\)

    Now we find \( \int_{-2}^{-1} f(x) \, dx \)

    \( \int_{-2}^{-1} \frac{x^2 + 6x + 5}{(x - 2)(x^2 + x + 1)} \, dx = 4ln(3) - 3ln(4)≈ 0.2356 \)

    Exercise &&1&& (&&1&& Question)

    Resolve into partial fractions:

    \( \frac{3x - 2}{x^2 - 4} \)

    1
    Submit

    Exercise &&2&& (&&1&& Question)

    Resolve into partial fractions:

    \( \frac{4x + 7}{x^2+x- 6} \)

    2
    Submit

    Exercise &&3&& (&&1&& Question)

    Resolve into partial fractions:

    \( \frac{2x - 2}{(x+1)(x-2)^2} \)

    3
    Submit

    Exercise &&4&& (&&1&& Question)

    Resolve the expression into partial fractions:

    \( \frac{3x^2 - 4x - 2}{(x - 1)(x - 2)} \)

    4
    Submit

    Exercise &&5&& (&&1&& Question)

    Express \( \frac{3x + 1}{(x + 1)(x^2 + x + 1)} \) in partial fractions.

    5
    Submit

    Exercise &&6&& (&&1&& Question)

    Find an antiderivative of: \( \frac{x^3 + 3}{x^2 - x} \)

    6
    Submit

    Exercise &&7&& (&&1&& Question)

    Decompose into partial fraction and find antiderivatives of: \( \frac{x + 3}{x^2 - 3x + 2} \)

    7
    Submit

    Exercise &&8&& (&&1&& Question)

    Evaluate the following: \( \int_{0}^{1} \frac{1 - 4x}{3 + x - 2x^2} \, dx \)

    8
    Submit

    Exercise &&9&& (&&1&& Question)

    Evaluate the following: \( \int_{0}^{1} \frac{x^2 - 1}{x^2 + 1} \, dx \)

    9
    Submit