A guide to find the integral of a rational function by decomposing it in partial fractions

The calculation of the integral of some functions implies the use of some ingenious techniques. In this post I am going to expose a method to find the integral of a rational function. It consists in decomposing the rational function into partial functions easier to integrate. This method is called "integration by partial decomposition".

In order to apply the techniques we make the following considerations:

1. The degree of the numerator must be one degree less than that of the denominator
2. The denominator must be decomposed into a product of factors. The number of factors will determine the number of partial fractions. Example: the function 2x + 3/(x+1)(x+2) can be decomposed as the sum of A/x+1 + B/x+2.
3. If the denominator is not decomposed into a product of factors it has to be so. Example: the function f(x) =  3x + 1/x² + 5x + 6 cannot be decomposed unless we factor it in a product of factors. Factoring the denominator as (x+2)(x+3) the function can be decomposed as: 3x + 1/(x+2)(x+3) = A/x+2 + B/x+3
4.  If the denominator is a power like (x + a)ⁿ then a rational function like N/(x + a)ⁿ can be written as: N/(x + a)ⁿ = A₁/x + a + A₂/(x+a)² + A₃/(x + a)³+.....An/(x+a)ⁿ

Example 1  Find the integral of the function f(x) = 2x+3/(x+1)(x+2)

The denominator is decomposed into 2 factors.  The fraction can be decomposed  into 2 partial fractions as follows:

2x+3/(x+1)(x+2) = A/x+1 + B/x+2

The goal here is to determine A and B. In order to do that, we have to go through a number of steps

1) Making the equality above identically equal. That will happen when the denominator of the fraction on the second side of the equation is equal to to the denominator of the fraction on the first side of the equation. The numerator of the fraction on the second side is written in the same way as the numerator of the fraction on the first side of the equation
2) Set a system of 2 equations containing A and B in order to determine these unknown
3) Substitute A and B in the expression: 2x+3/(x+1)(x+2) = A/x+1 + B/x+2

Let's do step 1:

2x+3/(x+1)(x+2) = A(x+2) + B(x + 1)/ (x+1)(x+2)

2x+3/(x+1)(x+2) = Ax + 2A + Bx + B/ (x + 1) (x + 2)

 2x+3/(x+1)(x+2) = (A + B)x + 2A + B/(x + 1) (x + 2

Let's simplify by (x + 1) (x + 2):

2x + 3 = (A + B)x + 2A + B

Now we can go to step 2:

By identification we have: A + B = 2 (1)  3 (2) 2A + B = 3

Let's solve this system of the 2 equarions by multiplying the first equarion by -2 and adding the new equation to the second equation

-2A-2 B = -4
2A + B = 3

We then obtain: -B = -1 B = 1

Let's substitute B in the second equation
:2A + 1 = 3
2A = 2  A = 1

Let's go to the final step 3:

Let's substitute A and B in the expression developed above:

2x + 3/(x+1)(x+2) = 1/x + 1 + 1/x+2

Now we can calculate the integral of 2x + 3/ (x + 1) (x + 2):

∫ (2x + 3/(x+1)(x+2) = ∫(1/x + 1 + 1/x + 2)dx

                                = ∫1/x + 1dx +  ∫1/x + 2 dx

                                = ln❘x + 1❘ + ln❘x + 2❘ + C

Example 2. Find the integral of the function f(x) =  3x + 1/x² + 5x + 6

Let's factor the denominator: x² + 5x + 6 = (x + 2) (x + 3)

Now the function can be written as:  f(x) =  3x + 1/ (x + 2) (x + 3)

Let's decompose the function in partial fractions:

3x + 1/ (x + 2) (x + 3) = A/x + 2 + B/x + 3

Step 1

In order to determine A and B let's do the first step which consists in making the expression above identically equal:

3x + 1/ (x + 2) (x + 3) = A(x + 3) + B(x + 2)/ (x + 2) (x + 3)

The equality of these two fractions implies the equality of the denominators and that of the numerators. The denominators being equal, let's equalize the numerators.

3x + 1 = A(x + 3) + B(x + 2)

3x + 1 = Ax + 3A + Bx + 2B

  3x + 1 = (A + B)x + 3A + 2B

Now we go to step 2. The equality is true if the coefficients of x and the constant terms are equal:

Step 2

A + B = 3 (1)  3A + 2B = 1 (2)

Let's multiply the first equation by -3 and add the new equation to the second equation
-3A - 3B = -6 (3)
  3A + 2B = 1 (2)

B = -5

Let's substitute B in the first equation

A-5 = 3 A = 8

Step 3

Let's substitute A and B in the decomposed function above:

    3x + 1/ (x + 2) (x + 3) = 8/x + 3 - 5/x + 2

Let's integrate both sides
:
∫3x + 1/ (x + 2) (x + 3)dx = ∫(8/x + 3 - 5/x + 2)dx =  ∫8/x + 3dx - ∫5/x + 2)dx = 8∫1/x + 3dx -5∫1/x + 2dx = 8ln❘x + 3❘-5ln❘x + 2❘ + C

From the example above we can set the general rule:

1) Decompose the rational function into an expression containing unknowns to determine: A, B, C, etc.
2) Make the equation identically equal
3) Set a system of equations that allow to determine the unknowns
4) When the unknowns are determined. substitute them in the function theoretically decomposed.
5) Calculate the integral 

Practice: Integrate the function: f(x) = 2x-3/(3x + 1)²

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The art of mastering the integration by parts

The integration parts allows to calculate the integral of the product of some functions. Let's derive a formula for it:

The formula for the derivative of a product is (uv)' = u'v + uv'

Then uv' = (uv)'-u'v

Let's integrate the 2 sides;

∫uv'dx = ∫(uv)'dx-∫u'vdx

∫u.(dv/dx).dx = uv-∫(du/dx).vdx

Let's simplify:

∫udv = uv-∫vdu

Example 1. Evaluate ∫xsinxdx;

Let's apply the formula: ∫udv = uv-∫vdu

Let's choose u = x and dv = sinxdx. Then du = dx  v = -cosx

Let's substitute u, v and dv in the formula:

∫xsinxdx = (x)(-cosx)- ∫(-cosxdx)dx
              = -xcosx + ∫cosxdx
              =  -xcosx + sinxdx + C

Note. Using the integration by parts can be tricky in choosing u and dv. To avoid bumping into a new integral that's not easy to calculate, choose u to have a derivative easier to calculate. Choose dv into an integral easy to calculate. In general you can use the following guide:
1) Choose u to be the part whose derivative is simpler than u. Use dv as the remaining term.
2) Choose dv to be the portion whose integral can be calculated using the basic formula. Choose u as the remaining term.

Application:  In example 1, choose u and dv differently to see if the integral is easy to calculate.

Example 2. Evaluate ∫xe^xdx ( read x exponential x dx). The sign ^ is used to signify that x is the exponent of e.

We can choose u first and choose dv as the remaining term. If so the derivative of u should be simpler than u. We can choose dv first so that the derivative can be calculated using a basic formula and choose u as the remaining portion of the integral.

Let's choose u = x and dv = e^x dx. Then du = dx and v = e^x.
Applying the formula:  ∫xe^xdx = xe^x-∫ e^x.dx
                                                   = xe^x- e^x + C
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An approach to calculate the area under curve: the definite integral

Let's suppose we have a function f and we want to calculate the area under its curve. It's not an easy thing to do. First, it is difficult to find the exact area under the curve. The easiest method would be to divide the area under the curve into a certain number of rectangles and to add these rectangles in order to get an approximation. The approximation is better if we divide the area in smaller and smaller rectangles.

Let's have a function f positive in an interval [x₀, x₄] and find the area under the curve of f between x₀ and x₄. The method consists of approximating the value of an area by dividing the area under the curve of (f) by a certain number of rectangles. Let's choose to divide the area under the curve into 4 rectangles of equal width Δx (delta x)


Let's name the area S(4). Then:

Let's divide S by 5 rectangles we have:

S(5) is greater than S(4) and represents better the area of the rectangle.
 .
The following simulation represents numerical values of the area S for a corresponding number of rectangles.

Each time we divide the area by a greater number the corresponding area approximates a greater number which is a better approximation. This fact is illustrated in the following figure:


In the figure above for n = 1 the approximated area is 14.62. For n = 2 the approximation is 51.20. For n = 4 the approximation is 63.02. For n = 8, the approximation 67.32. For n = 16 the approximation is 69.07. For n = 32 the approximation is 69.84. For n = 64 the approximation is 70.20.

If we divide the area by n rectangles the area S(n) can be calculated by:

S(n) can be written as S(n) = Σ f(xi)𝚫x with i varying from 1 to n or:

As n comes to an extremely greater number the area approximates a greater number. The approximation becomes better and better and comes close to a greater number. This greatest number is the best approximation for the area of the rectangle. It is the limit of all the approximations when the number of rectangles n becomes extremely high. This limit is the limit of the function S(n). We write:

This value represents the integral of the function f between x0 and x1

We write:

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Rules of integration for combination of functions

If f and g are two integrable functions and C a constant, then we can derive the rules for the integration of the sum, difference of two functions as well as the multiplication of a function by a constant.

Integral of the sum of two functions

∫[f(x)+g(x)]dx = ∫f(x)dx + ∫g(x)dx

The integral of the sum of two functions is equal to the sum of the integrals of these functions. This rule is valid for the integral of the indefinite sum of functions. The integral of the sum of three functions is equal to the sum of the integrals of these three functions. The integral of the sum of four functions is equal to the sum of the integrals of these four functions. The integral of the sum of n functions is equal to the sum of the integrals of these n functions.

Integral of the difference of two functions

∫[f(x) - g(x)]dx = ∫f(x)dx - ∫g(x)dx

Integral of the product of a constant C by a function

∫(Cf(x)dx) = C∫(x)dx.

Integration by substitution

Sometimes it is difficult to integrate a function using the simple formulas of integrals. We have to introduce a second variable to make the integration easier.

Example 1

Calculate ∫(x+1)⁵dx

If it was the integral of x⁵ it would be easy to calculate. We can introduce another variable to make the integral easy to calculate. If we make x + 1 equal to u we will just have to calculate the integral of u. So let's write u = x+1

Then our integral becomes ∫u⁵dx. We calculate the integral as if it was the integral of x⁵.

 ∫u⁵dx = u⁵⁺¹/5+1 + C = 1/6 u⁶ + C

Let's substitute u we get  ∫u⁵dx = 1/6 (x+1)⁶ + C

Example 2

Calculate ∫√4x+3dx

Let's write the expression as a power:

 ∫√4x+3dx = ∫(4x+3)¹/² dx

Let's write u = 4x+3 du = 4dx then dx = du/4. Let's substitute 4x+3 and dx

∫√4x+3dx = ∫u¹/².du/4 = 1/4∫u¹/²du = 1/4(u¹/²+¹/¹/²+¹) + C = 1/4(u³/²/3/2) + C = (1/4)(2/3)u³/² + C = 1/6u³/² + C = 1/6√(4x+3)³ + C

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Using derivative rules to find the integrals of some basic functions

Integral of the power function

Let f be f(x) = xⁿ its antiderivative is F(x) = 1/n+1x^n+1+C (Read F(x) equals  1 over n+1 times x exponent n+1 plus the constant of integration C) . You can rewrite the expression by utilizing a horizontal bar for the fraction 1 over n+1. You can also rewrite the expression x^n+1 by putting n+1 besides x and above. The rewriting will make the expression clearer.

 The antiderivative of f(x) being the indefinite integral, we can write ∫f(x)dx =  1/n+1.x^n+1+C (Read sum f(x) dx equals 1 over n+1 times  x exponent n+1 plus the constant of integration C).

In order to prove this, we have to demonstrate that the derivative of  1/n+1.x^n+1+C is xⁿ.

Let's calculate the derivative of F(x) = 1/n+1.x^n+1+C 
F'(x) = n+1/n+1x^n+1-1 (Read n+1 over n+1 times x exponent n+1-1. By simplifying and reducing F'(x) = xⁿ

Integral of the inverse function

The integral of the inverse function f(x) = 1/x is ∫1/xdx = lnx +C.

Proof

The derivative of F(x) = lnx is F'(x) = 1/x

Integral of the exponential function

The integral of the exponential function f(x) = e^x is ∫e^xdx = e^x + C. The term e^x reads e exponential x.

Proof

The derivative of F(x) = e^x is F'(x) = e^x.

Rules of integrable functions

If f and g are integrable functions and C is a constant, then

∫[f(x)+g(x)] = ∫f(x)dx+∫g(x)dx

∫[f(x)-g(x)] = ∫f(x)dx-∫g(x)dx

∫kf(x)dx = k∫f(x)dx

Example

Solution

Using the basic rules of integration we have:

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The formula of the antiderivative of a function simply explained

Let's consider the function f(x) = x² +3 its derivative is f '(x) = 2x. The function f '(x) = 2x is the derivative of the function f(x) = x² + 3. The function f(x) = x² +3 is the antiderivative of the function
f '(x) = 2x. The antiderivative is the function from which the derivative was calculated.

If we consider the function f(x) = 2x it can be considered as the derivative of another function F. This function which is F(x) = x² + 3 is the antiderivative of the function f. Actually, there are several functions whose derivative is f(x) = 2x. Some examples of these functions are F(x) = x²+1, F(x) = x²-3, F(x) = x² + 6. There is a bunch of functions whose derivative is f '(x) = 2x. We can't list them all. For this reason we represent all them by the function F(x) = x²+C.

Notation and notion of indefinite integral

The process of finding the antiderivative of a function is called antidifferentiation or integration. If F is the antiderivative of a function f we write ∫f(x)dx = F(x) +C. The left side of this equation is called integral of f(x). The function f(x) is called the integrand. The letter C is the constant of integration. The symbol dx means that the function is integrated with respect to x.

Formula to calculate the antiderivative of the function power

Let's retake the function f(x) = 2x. Its antiderivative is F(x) = x² + C or F(x) = 2/2x² + C.  Let's consider another function g(x) = 3x. Its antiderivative can be G(x) = 3/2x²+C. For the function h(x) = x² its antiderivative is H(x) = 1/3x³+C. For the function i(x) =5 x³ its antiderivative is I(x) = 5/4x⁴ + C

By observing each of these functions and their antiderivative, two points draw our attention.

1) The exponent of x in the antiderivative function is one degree more than the exponent of x in the given function.
2) The coefficient of x in the antiderivative function is obtained by dividing the coefficient of x in the given function by the exponent of x in the antiderivative function.

Let's come back to the examples above. In the function antiderivative F(x) the exponent of x is 2 and the exponent of x in the function f is 1. The coefficient of x 2/2 in the function F(x) is equal to the coefficient 2 of x in the function f(x) divided by the exponent 2 of x in the antiderivative function.

 In the function antiderivative G(x) the exponent of x is 2 and the exponent of x in the function g(x) is 1. The coefficient 3/2 of x in the function G(x) is equal to the coefficient 3 of x in the function f(x)   divided by the exponent 2 of x in the function G(x). The same is true for the functions H(x) and I(x).

From these observations, we can deduct a rule. In order to find the antiderivative of f((x) = xⁿ add 1 to the exponent and divide the coefficient by the new exponent,

The formula that generalizes this procedure is as follows:

If f(x) = kxⁿ its derivative is given by F(x) = k/n+1xⁿ⁺¹ + C.

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Formal definition of limits

The formal definition of limits is mostly skipped in teaching about limits. The intuitive notion is the most taught. The formal definition of limits is interesting and is derived from the intuitive notion.

The intuitive definition says that the limit of a function f is L when x approaches a real number "a" if f becomes closer and closer to "L". We write lim f(x) = L when x➡ a  .

 When x comes closer and closer to a number "a" both to the left and right of that number, x gets values in the neighborhood of a. The variable x in approaching to "a" comes to a certain distance to "a" both from the left and the right. As this distance is very small we call it δ . The variable "x' takes values in the interval ⦐a-ẟ  a+ẟ[. The set of values "x" are translated into the equation ᥣx-aᥣ<ε.

The independent variable f(x) comes to a certain distance of "L" both from the right and the left. As this distance is very small we call it δ. The independent variable f(x) takes values in the interval ]f(x)-ε  f(x)+ε[.

When x approaches "a" f(x) approaches  "L". In other words you give me an ϵ such that ❙f(x)-a❙<ε I will find you a δ that satisfies the equation ❙x-a❙<δ.


The formal definition of limits is stated:

Method

You give me the ε from the equation ❘f(x)-L❘<ε. I'll transform this equation in ❘x-xindice0❘<δ in order to find δ.

However when the inequality ❘f(x)-L❘<ϵ is transformed in a second degree inequality the process becomes more complicated. Follow the process of solving the example below to solve similar examples..

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