Integration by Partial Fractions

THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS

All of the following problems use the method of integration by partial fractions. This method is based on the simple concept of adding fractions by getting a common denominator. For example,

$\displaystyle{ {1 \over 2} + {1 \over 3} } = \displaystyle{ {3 \over 6} + {2 \over 6} } = \displaystyle{ 5 \over 6 }$

so that we can now say that a partial fractions decomposition for $\ \displaystyle{ 5 \over 6 } \$ is

$\displaystyle{5 \over 6} = \displaystyle{ {1 \over 2} + {1 \over 3} }$ .

This concept can also be used with functions of . For example,

$\displaystyle{ {2 \over x+1} - {1 \over x} } = \displaystyle{ {2 \over x+1}\Big({x \over x}\Big) - {1 \over x}\Big({x+1 \over x+1}\Big) }$

$= \displaystyle{ {2x \over x(x+1) } - {x+1 \over x(x+1) } }$

$= \displaystyle{ 2x-x-1 \over x^2+x }$

$= \displaystyle{ x-1 \over x^2+x }$

so that we can now say that a partial fractions decomposition for $\ \displaystyle{ x-1 \over x^2+x } \$ is

$\displaystyle{ x-1 \over x^2+x } = \displaystyle{ {2 \over x+1} - {1 \over x} }$ .

Of course, what we would like to be able to do is find a partial fractions decomposition for a given function. For example, what would be a partial fractions decomposition for $\ \displaystyle{ 6 \over x^2-1 }$ ? Begin by factoring the denominator, getting

$\displaystyle{ 6 \over x^2-1 } = \displaystyle{ 6 \over (x-1)(x+1) }$ .

Now ASSUME that there are constants and so that

( equation (*) ) $\ \ \ \ \ \ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A \over x-1 } + { B \over x+1 } }$ .

It can be shown that such constants always exist for the rational function $\ \displaystyle{ p(x) \over q(x) } \$ if the following two conditions are met. They are

1. both $\ p(x) \$ and $\ q(x) \$ are polynomials (constants together with positive integer powers of only)

and

2. the degree (highest power of ) of $\ p(x) \$ is smaller than the degree of $\ q(x) \$ .

Let's proceed. So far we have

$\ \ \ \ \ \ \displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ { A \over x-1 } + { B \over x+1 } }$

(Get a common denominator and add the fractions.)

$= \displaystyle{ {A \over x-1}\Big({x+1 \over x+1}\Big) + {B \over x+1}\Big({x-1 \over x-1}\Big) }$

$= \displaystyle{ {A(x+1) \over (x-1)(x+1)} + {B(x-1) \over (x-1)(x+1)} }$

$= \displaystyle{ A(x+1) + B(x-1) \over (x-1)(x+1) }$ ,

that is,

$\displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ A(x+1) + B(x-1) \over (x-1)(x+1) }$ .

Since the fractions in the above equation have the same denominators, it follows that their numerators must be equal. Thus,

( equation (**) ) $\ \ \ \ \ \ 6 = A(x+1) + B(x-1)$ .

The right-hand side of this equation can be considered a function of which is equal to 6 for all values of . In particular, it must also be true for specific values of . For example, if we choose to

let $\ \ \ x=1$ ,

then

so that

If we choose to

let $\ \ \ x=-1$ ,

then

so that

We can now say that a partial fractions decomposition for $\ \displaystyle{ 6 \over (x-1)(x+1) \ }$ is

$\displaystyle{ 6 \over (x-1)(x+1) } = \displaystyle{ {3 \over x-1} + {-3 \over x+1} }$ .

It should be noted that $\ \ \ x=1$ and $\ \ \ x=-1$ were chosen for use in equation (**) for their convenience of ``zeroing out" terms in the equation. However, any other two choices for will lead to the exact same values for and (after solving two equations with two unknowns). Try it. After getting familiar with this process, in order to save some time, get in the habit of going from the step at equation (*) directly to the step at equation (**). Here is another important point to consider when applying the method of partial fractions to the rational function $\ \displaystyle{ p(x) \over q(x) } \$ . If the degree (highest power) of $\ p(x) \$ is equal to or greater than the degree of $\ q(x) \$ , then you must use polynomial division in order to rewrite the given rational function as the sum of a polynomial and a new rational function satisfying condition 2 above. For example, polynomial division leads to

$\displaystyle{ 4x^3-3x+5 \over x^2-2x } = 4x+8 +\displaystyle{ 13x+5 \over x^2-2x }$ ,

where the rational function on the right-hand side of the equation satisfies condition 2. There are other points to consider. Recall that the complex number $i = \sqrt{ -1 }$ so that and . In addition, if two complex numbers are equal, then their real and complex components are equal. That is, if

then

$a = c \ \ \$ and $\ \ \ b = d$ .

Now let's do another example. Find a partial fractions decomposition for $\ \displaystyle{ 2x-3 \over x^3+x } \$ . Begin by factoring the denominator, getting

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

Now ASSUME that there are constants and so that

( equation (*) ) $\ \ \ \ \ \ \displaystyle{ 2x-3 \over x(x^2+1) } = \displaystyle{ { A \over x } + { Bx+C \over x^2+1 } }$ .

Since $\ x^2+1 \$ is an irreducible quadratic expression, assuming only that

$\displaystyle{ 2x-3 \over x(x^2+1) } = \displaystyle{ { A \over x } + { B \over x^2+1 } }$

is NOT GENERAL ENOUGH and will not always lead to a correct partial fractions decomposition. Continuing, we have

$\ \ \ \ \ \ \displaystyle{ 2x-3 \over x(x^2+1) } = \displaystyle{ { A \over x } + { Bx+C \over x^2+1 } }$

(Get a common denominator and add the fractions.)

$= \displaystyle{ {A \over x}\Big({x^2+1 \over x^2+1}\Big) + {Bx+C \over x^2+1}\Big({x \over x}\Big) }$

$= \displaystyle{ {A(x^2+1) \over x(x^2+1)} + {(Bx+C)x \over x(x^2+1) } }$

$= \displaystyle{ A(x^2+1) + (Bx+C)x \over x(x^2+1) }$ ,

that is,

$\displaystyle{ 2x-3 \over x(x^2+1) } = \displaystyle{ A(x^2+1) + (Bx+C)x \over x(x^2+1) }$ .

Since the fractions in the above equation have the same denominators, it follows that their numerators must be equal. Thus,

( equation (**) ) $\ \ \ \ \ \ 2x-3 = A(x^2+1) + (Bx+C)x$ .

This equation can be considered two functions of which are equal to each other for all values of . In particular, it must also be true for specific values of . For example, if we ``conveniently" choose to

let $\ \ \ x=0$ ,

then

so that

If we choose to

let $\ \ \ x=i$ ,

then

so that

$B = 3 \ \ \$ and $\ \ \ C = 2$ .

We can now say that a partial fractions decomposition for $\ \displaystyle{ 2x-3 \over x(x^2+1) } \$ is

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

If you choose to NOT use complex numbers to solve for the unknown constants in the previous example, using TWO other real values of instead of $\ \ \ x=i$ will lead to the exact same values for and . There is one final case to consider. How should repeated factors in the denominator be handled ? The following example illustrates the partial fractions decomposition of a rational function, where the linear factor $\ x+1 \$ is repeated three times and the irreducible quadratic factor $\ x^2+4 \$ is repeated twice. Thus,

$\displaystyle{ 2x^3+5x-1 \over (x+1)^3(x^2+1)^2 } = \displaystyle{ { A \over... ...^2 } + { C \over (x+1)^3 } + { Dx+E \over x^2+1 } + { Fx+G \over (x^2+1)^2 } }$ .

We will assume familiarity with the following rules of differentiation.

a.) $D (\sin x) = \cos x$
b.) $D (\tan x) = \sec^2 x$
c.)

We will also assume knowledge of the following well-known, basic indefinite integral formulas :

$\displaystyle{ \int x^n \,dx } = \displaystyle{ {x^{n+1} \over n+1 } + C }$ , where is a constant $(n \ne -1 )$
$\displaystyle{ \int { 1 \over x } \,dx } = \displaystyle{ \ln \vert x\vert + C }$
$\displaystyle{ \int { 1 \over x^2+a^2 } \,dx } = \displaystyle{ {1 \over a} \arctan {x \over a} + C }$
$\displaystyle{ \int k f(x) \,dx } = k \displaystyle{ \int f(x) \,dx }$ , where is a constant
$\displaystyle{ \int ( f(x) \pm g(x) ) \,dx } = \displaystyle{ \int f(x) \,dx } \pm \displaystyle{ \int g(x) \,dx }$

Most of the following problems are average. A few are challenging. It is assumed that you are proficient with the method of u-substitution. Make careful and precise use of the differential notation

and

and be careful when arithmetically and algebraically simplifying expressions. Solutions to all of the following problems will use the equation (*) to equation (**) shortcut illustrated in the two examples given above.

PROBLEM 1 : Integrate $\displaystyle{ \int { 1 \over x^2-4 } \,dx }$ .
Click HERE to see a detailed solution to problem 1.
PROBLEM 2 : Integrate $\displaystyle{ \int { 2x+3 \over x^2-9 } \,dx }$ .
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PROBLEM 3 : Integrate $\displaystyle{ \int { 2-x \over x^2+5x } \,dx }$ .
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PROBLEM 4 : Integrate $\displaystyle{ \int { x^2-1 \over x^2-16 } \,dx }$ .
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PROBLEM 5 : Integrate $\displaystyle{ \int { x^4+x^3+x^2+1 \over x^2+x-2 } \,dx }$ .
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PROBLEM 6 : Integrate $\displaystyle{ \int { x^2+x-1 \over x(x^2-1) } \,dx }$ .
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PROBLEM 7 : Integrate $\displaystyle{ \int { x+7 \over x^2(x+2) } \,dx }$ .
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PROBLEM 8 : Integrate $\displaystyle{ \int { x^5+1 \over x^3(x+2) } \,dx }$ .
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PROBLEM 9 : Integrate $\displaystyle{ \int { x^2-x+1 \over (x+1)^3 } \,dx }$ .
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PROBLEM 10 : Integrate $\displaystyle{ \int { x^3+4 \over (x^2-1)(x^2+3x+2) } \,dx }$ .
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PROBLEM 11 : Integrate $\displaystyle{ \int { x^3+2x-1 \over (x^2-x-2)^2 } \,dx }$ .
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PROBLEM 12 : Integrate $\displaystyle{ \int { \sec^2 x \over \tan^3 x - \tan^2 x } \,dx }$ .
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PROBLEM 13 : Integrate $\displaystyle{ \int{ x^3+8 \over (x^2-1)(x-2) } \,dx }$ .
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PROBLEM 14 : Integrate $\displaystyle{ \int { e^x \over (e^x-1)(e^x+3) } \,dx }$ .
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PROBLEM 15 : Integrate $\displaystyle{ \int { 1 \over e^x+1 } \,dx }$ .
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PROBLEM 16 : Integrate $\displaystyle{ \int { 3-x \over x(x^2+1) } \,dx }$ .
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PROBLEM 17 : Integrate $\displaystyle{ \int { 3x+1 \over x^2(x^2+25) } \,dx }$ .
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PROBLEM 18 : Integrate $\displaystyle{ \int { 1 \over x^4-16 } \,dx }$ .
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PROBLEM 19 : Integrate $\displaystyle{ \int { \cos x \over \sin^3 x + \sin x } \,dx }$ .
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PROBLEM 20 :Integrate $\displaystyle{ \int{ 1 \over x^4+4 } \,dx }$ .
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