2.4: Partial Fractions
- Page ID
- 128835
- Integrate a rational function using the method of partial fractions.
- Recognize simple linear factors in a rational function.
- Recognize repeated linear factors in a rational function.
- Recognize quadratic factors in a rational function.
We have seen some techniques that allow us to integrate specific rational functions. For example, we know that
\[ \int \dfrac{du}{u}=\ln |u|+C \nonumber \]
and
\[ \int \dfrac{du}{u^2+a^2}=\dfrac{1}{a}\tan^{−1} \left(\dfrac{u}{a}\right)+C.\nonumber \]
However, we do not yet have a technique that allows us to tackle arbitrary quotients of this type. Thus, it is not immediately obvious how to go about evaluating
\[ \int \dfrac{3x}{x^2−x−2}\,dx.\nonumber \]
However, we know from material previously developed that
\[ \int \left(\dfrac{1}{x+1}+\dfrac{2}{x−2}\right)\,dx=\ln |x+1|+2\ln |x−2|+C.\nonumber \]
In fact, by getting a common denominator, we see that
\[ \dfrac{1}{x+1}+\dfrac{2}{x−2}=\dfrac{3x}{x^2−x−2}.\nonumber \]
Consequently,
\[ \int \dfrac{3x}{x^2−x−2}\,dx=\int \left(\dfrac{1}{x+1}+\dfrac{2}{x−2}\right)\,dx.\nonumber \]
In this section, we examine the method of Partial Fraction Decomposition, which allows us to decompose rational functions into sums of simpler, more easily integrated rational functions. Using this method, we can rewrite an expression such as:
\[ \dfrac{3x}{x^2−x−2}\nonumber \]
as an expression such as
\[ \dfrac{1}{x+1}+\dfrac{2}{x−2}.\nonumber \]
The key to the method of partial fraction decomposition is being able to anticipate the form that the decomposition of a rational function will take. As we shall see, this form is both predictable and highly dependent on the factorization of the denominator of the rational function. It is also extremely important to keep in mind that partial fraction decomposition can be applied to a rational function \( \frac{P(x)}{Q(x)}\) only if \( \deg(P(x)) \lt \deg(Q(x))\). In the case when \( \deg(P(x)) \geq \deg(Q(x))\), we must first perform long division1 to rewrite the quotient \( \frac{P(x)}{Q(x)}\) in the form \( A(x)+\frac{R(x)}{Q(x)}\), where \( \deg(R(x)) \lt \deg(Q(x))\). We then do a partial fraction decomposition on \( \frac{R(x)}{Q(x)}\). The following example, although not requiring partial fraction decomposition, illustrates our approach to integrals of rational functions of the form \( \int \frac{P(x)}{Q(x)}\,dx\), where \( \deg(P(x)) \geq \deg(Q(x)).\)
Evaluate
\[ \int \dfrac{x^2+3x+5}{x+1}\,dx. \nonumber \]
Solution
Since \( \deg(x^2+3x+5) \geq \deg(x+1),\) we perform long division to obtain
\[ \dfrac{x^2+3x+5}{x+1}=x+2+\dfrac{3}{x+1}. \nonumber \]
Thus,
\[ \int \dfrac{x^2+3x+5}{x+1}\,dx=\int \left(x+2+\dfrac{3}{x+1}\right)\,dx=\dfrac{1}{2}x^2+2x+3\ln |x+1|+C. \nonumber \]
Evaluate
\[ \int \dfrac{x−3}{x+2}\,dx. \nonumber \]
- Hint
-
Use long division to obtain \( \frac{x−3}{x+2}=1−\frac{5}{x+2}\).
- Answer
-
\[ x−5\ln |x+2|+C \nonumber \]
To integrate \(\displaystyle \int \frac{P(x)}{Q(x)}\,dx\), where \( \deg(P(x)) \lt \deg(Q(x))\), we must begin by factoring \( Q(x)\).
Nonrepeated Linear Factors
If \( Q(x)\) can be factored as \( (a_1x+b_1)(a_2x+b_2) \ldots (a_nx+b_n)\), where each linear factor is distinct, then it is possible to find constants \( A_1,A_2, \ldots A_n\) satisfying the following.2
\[ \dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+ \cdots +\dfrac{A_n}{a_nx+b_n}. \label{eq:7.4.1} \]
In this next example, we see how to use partial fractions to integrate a rational function of this type.
Evaluate
\[\int \dfrac{3x+2}{x^3−x^2−2x}\,dx. \nonumber \]
Solution
Since \( \deg(3x+2) \lt \deg(x^3−x^2−2x)\), we begin by factoring the denominator of \( \frac{3x+2}{x^3−x^2−2x}\).
We can see that \( x^3−x^2−2x=x(x−2)(x+1)\). Thus, there are constants \(A\), \(B\), and \( C\) satisfying Equation \ref{eq:7.4.1} such that
\[ \dfrac{3x+2}{x(x−2)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x−2}+\dfrac{C}{x+1}. \nonumber \]
We must now find these constants. To do so, we begin by multiplying both sides of this equation by the least common denominator: \( x(x - 2)(x + 1) \). Thus,
\[ 3x+2=A(x−2)(x+1)+Bx(x+1)+Cx(x−2).\label{Ex2Numerator} \]
There are two different strategies for finding the coefficients \(A\), \(B\), and \(C\). We refer to these as the method of equating coefficients and the method of strategic substitution.
Strategy One: Method of Equating Coefficients
Rewrite Equation \(\ref{Ex2Numerator}\) in the form
\[ 3x+2=(A+B+C)x^2+(−A+B−2C)x+(−2A). \nonumber \]
Equating coefficients produces the system of equations
\[\begin{array}{rcl}
A+B+C & = & 0 \\
−A+B−2C & = & 3 \\
−2A & = & 2. \\
\end{array} \nonumber \]
To solve this system, we first observe that \( −2A=2 \implies A=−1.\) Substituting this value into the first two equations gives us the system
\[ \begin{array}{rcl}
B+C & = & 1 \\
B−2C & = & 2. \\
\end{array} \nonumber \]
Multiplying the second equation by \( −1\) and adding the resulting equation to the first produces \( −3C=1,\) which in turn implies that \( C=−\frac{1}{3}\). Substituting this value into the equation \( B+C=1\) yields \( B=\frac{4}{3}\). Thus, solving these equations yields \( A=−1\), \(B=\frac{4}{3}\), and \( C=−\frac{1}{3}\).
It is important to note that the system produced by this method is consistent if and only if we have set up the decomposition correctly. If the system is inconsistent, there is an error in our decomposition.
Strategy Two: Method of Strategic Substitution
The method of strategic substitution is also based on the assumption that we have set up the decomposition correctly. If the decomposition is set up correctly, then there must be values of \( A\), \(B\), and \( C\) that satisfy Equation \(\ref{Ex2Numerator}\) for all values of \( x\). That is, this equation must be true for any value of \( x\) we care to substitute into it. Therefore, by choosing values of \( x\) carefully and substituting them into the equation, we may find \( A\), \(B\), and \( C\) easily.
For example, if we substitute \( x=0\), the equation reduces to \( 2=A(−2)(1)\). Solving for \( A\) yields \( A=−1\). Next, by substituting \( x=2\), the equation reduces to \( 8=B(2)(3)\), or equivalently \( B= \frac{4}{3}\). Last, we substitute \( x=−1\) into the equation and obtain \( −1=C(−1)(−3).\) Solving, we have \( C=−\frac{1}{3}\).
It is important to keep in mind that if we attempt to use this method with a decomposition that has not been set up correctly, we are still able to find values for the constants, but these constants are meaningless. If we do opt to use the method of strategic substitution, then it is a good idea to check the result by recombining the terms algebraically.
Now that we have the values of \( A\), \(B\), and \( C\), we rewrite the original integral:
\[ \int \dfrac{3x+2}{x^3−x^2−2x}\,dx=\int \left(−\dfrac{1}{x}+\dfrac{4}{3} \cdot \dfrac{1}{x−2}−\dfrac{1}{3} \cdot \dfrac{1}{x+1}\right)\,dx. \nonumber \]
Evaluating the integral gives us
\[ \int \dfrac{3x+2}{x^3−x^2−2x}\,dx=−\ln |x|+\dfrac{4}{3}\ln |x−2|−\dfrac{1}{3}\ln |x+1|+C. \nonumber \]
In the next example, we integrate a rational function in which the degree of the numerator is not less than the degree of the denominator.
Evaluate
\[\int \dfrac{x^2+3x+1}{x^2−4}\,dx.\nonumber \]
Solution
Since \( \deg(x^2+3x+1) \geq \deg(x^2−4)\), we must perform long division of polynomials. This results in
\[ \dfrac{x^2+3x+1}{x^2−4}=1+\dfrac{3x+5}{x^2−4}. \nonumber \]
Next, we perform partial fraction decomposition on \( \frac{3x+5}{x^2−4}=\frac{3x+5}{(x+2)(x−2)}\). We have
\[ \dfrac{3x+5}{(x−2)(x+2)}=\dfrac{A}{x−2}+\dfrac{B}{x+2}. \nonumber \]
Thus,
\[ 3x+5=A(x+2)+B(x−2). \nonumber \]
Solving for \( A\) and \( B\) using either method, we obtain \( A= \frac{11}{4}\) and \( B= \frac{1}{4}\).
Rewriting the original integral, we have
\[ \int \dfrac{x^2+3x+1}{x^2−4}\,dx=\int \left(1+\dfrac{11}{4} \cdot \dfrac{1}{x−2}+\dfrac{1}{4} \cdot \dfrac{1}{x+2}\right)\,dx. \nonumber \]
Evaluating the integral produces
\[ \int \dfrac{x^2+3x+1}{x^2−4}\,dx=x+\dfrac{11}{4}\ln |x−2|+\dfrac{1}{4}\ln |x+2|+C. \nonumber \]
As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. To do so, we must first convert the nonrational function to a rational function through a substitution.
Evaluate
\[ \int \dfrac{\cos x}{\sin^2x−\sin x}\,dx.\nonumber \]
Solution
Let’s begin by letting \( u=\sin x\). Consequently, \( du=\cos x\,dx\). After making these substitutions, we have
\[ \int \dfrac{\cos x}{\sin^2x−\sin x}\,dx=\int \dfrac{du}{u^2−u}=\int \dfrac{du}{u(u−1)}. \nonumber \]
Applying partial fraction decomposition to \(\frac{1}{u(u−1)}\) gives \( \frac{1}{u(u−1)}=−\frac{1}{u}+\frac{1}{u−1}\).
Thus,
\[ \int \dfrac{\cos x}{\sin^2x−\sin x}\,dx=−\ln |u|+\ln |u−1|+C=−\ln |\sin x|+\ln |\sin x−1|+C. \nonumber \]
Evaluate
\[ \int \dfrac{x+1}{(x+3)(x−2)}\,dx.\nonumber \]
- Hint
-
\[ \dfrac{x+1}{(x+3)(x−2)}=\dfrac{A}{x+3}+\dfrac{B}{x−2} \nonumber \]
- Answer
-
\[ \dfrac{2}{5}\ln |x+3|+\dfrac{3}{5}\ln |x−2|+C \nonumber \]
2 The proof that such constants exist is beyond the scope of this course.
Repeated Linear Factors
For some applications, we need to integrate rational expressions that have denominators with repeated linear factors - that is, rational functions with at least one factor of the form \( (ax+b)^n,\) where \( n\) is a positive integer greater than or equal to \( 2\). If the denominator contains the repeated linear factor \( (ax+b)^n\), then the decomposition must contain
\[ \dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+ \cdots +\dfrac{A_n}{(ax+b)^n}. \label{eq:7.4.2} \]
As we see in our next example, the basic technique used for solving for the coefficients is the same, but it requires more algebra to determine the numerators of the partial fractions.
Evaluate
\[ \int \dfrac{x−2}{(2x−1)^2(x−1)}\,dx.\nonumber \]
Solution
We have \( \deg(x−2) \lt \deg((2x−1)^2(x−1))\), so we can proceed with the decomposition. Since \( (2x−1)^2\) is a repeated linear factor, include
\[ \dfrac{A}{2x−1}+\dfrac{B}{(2x−1)^2} \nonumber \]
in the decomposition in Equation \ref{eq:7.4.2}. Thus,
\[ \dfrac{x−2}{(2x−1)^2(x−1)}=\dfrac{A}{2x−1}+\dfrac{B}{(2x−1)^2}+\dfrac{C}{x−1}. \nonumber \]
After multiplying both sides by the least common denominator, we have
\[ x−2=A(2x−1)(x−1)+B(x−1)+C(2x−1)^2. \label{Ex5Numerator} \]
We then use the method of equating coefficients to find the values of \( A, B,\) and \( C\).
\[ x−2=(2A+4C)x^2+(−3A+B−4C)x+(A−B+C). \nonumber \]
Equating coefficients yields \( 2A+4C=0\), \(−3A+B−4C=1\), and \( A−B+C=−2\). Solving this system yields \( A=2\), \(B=3\), and \( C=−1\).
Alternatively, we can use the method of strategic substitution. In this case, substituting \( x=1\) and \( x= \frac{1}{2}\) into Equation \(\ref{Ex5Numerator}\) easily produces the values \( B=3\) and \( C=−1\). At this point, it may seem that we have run out of good choices for \( x\); however, since we already have values for \( B\) and \( C\), we can substitute in these values and choose any value for \( x\) not previously used. The value \( x=0\) is a good option. In this case, we obtain the equation \( −2=A(−1)(−1)+3(−1)+(−1)(−1)^2\) or, equivalently, \( A=2.\)
Now that we have the values for \( A\), \(B\), and \( C\), we rewrite the original integral and evaluate it:
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{x−2}{(2x−1)^2(x−1)}\,dx & = & \displaystyle \int \left(\dfrac{2}{2x−1}+\dfrac{3}{(2x−1)^2}−\dfrac{1}{x−1}\right)\,dx \\
& = & \ln |2x−1|−\dfrac{3}{2(2x−1)}−\ln |x−1|+C. \\
\end{array}\nonumber \]
Set up the partial fraction decomposition for
\[ \int \dfrac{x+2}{(x+3)^3(x−4)^2}\,dx. \nonumber \]
(Do not solve for the coefficients or complete the integration.)
- Hint
-
Use the problem-solving method of Example \( \PageIndex{5}\) for guidance.
- Answer
-
\[ \dfrac{x+2}{(x+3)^3(x−4)^2}=\dfrac{A}{x+3}+\dfrac{B}{(x+3)^2}+\dfrac{C}{(x+3)^3}+\dfrac{D}{(x−4)}+\dfrac{E}{(x−4)^2} \nonumber \]
The General Method
Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let’s outline the basic method in the following problem-solving strategy.
To decompose the rational function \( \frac{P(x)}{Q(x)}\), use the following steps:
- Make sure that \( \deg(P(x)) \lt \deg(Q(x))\). If not, perform long division of polynomials.
- Factor \( Q(x)\) into the product of linear and irreducible quadratic factors.3
- Assuming that \( \deg(P(x)) \lt \deg(Q(x)\), the factors of \( Q(x)\) determine the form of the decomposition of \( \frac{P(x)}{Q(x)}\).
- If \( Q(x)\) can be factored as \( (a_1x+b_1)(a_2x+b_2) \ldots (a_nx+b_n)\), where each linear factor is distinct, then it is possible to find constants \( A_1,A_2,\ldots,A_n\) satisfying \[ \dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+ \cdots +\dfrac{A_n}{a_nx+b_n}. \nonumber \]
- If \( Q(x)\) contains the repeated linear factor \( (ax+b)^n\), then the decomposition must contain \[ \dfrac{A_1}{ax+b}+\dfrac{A_2}{(ax+b)^2}+ \cdots +\dfrac{A_n}{(ax+b)^n}. \nonumber \]
- For each irreducible quadratic factor \( ax^2+bx+c\) that \( Q(x)\) contains, the decomposition must include \[ \dfrac{Ax+B}{ax^2+bx+c}. \nonumber \]
- For each repeated irreducible quadratic factor \( (ax^2+bx+c)^n,\) the decomposition must include \[ \dfrac{A_1x+B_1}{ax^2+bx+c}+\dfrac{A_2x+B_2}{(ax^2+bx+c)^2}+ \cdots +\dfrac{A_nx+B_n}{(ax^2+bx+c)^n}. \nonumber \]
- After the appropriate decomposition is determined, solve for the constants.
- Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.
3 An irreducible quadratic is a quadratic that has no real zeros.
Irreducible Quadratic Factors
Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic \( ax^2+bx+c\) is irreducible if \( ax^2+bx+c=0\) has no real zeros - that is, if \( b^2−4ac \) (the discriminant) is less than \(0\).
Evaluate
\[ \int \dfrac{2x−3}{x^3+x}\,dx.\nonumber \]
Solution
Since \( \deg(2x−3) \lt \deg(x^3+x)\), factor the denominator and proceed with partial fraction decomposition. In this case, \( x^3+x=x(x^2+1)\) contains the irreducible quadratic factor \( x^2+1\). Therefore, we include \( \frac{Ax+B}{x^2+1}\) as part of the decomposition, along with \( \frac{C}{x}\) for the linear term \(x\). Thus, the decomposition has the form
\[ \dfrac{2x−3}{x(x^2+1)}=\dfrac{Ax+B}{x^2+1}+\dfrac{C}{x}.\nonumber \]
After multiplying both sides by the least common denominator, we obtain the equation
\[ 2x−3=(Ax+B)x+C(x^2+1).\nonumber \]
Solving for \( A\), \(B\), and \( C\), we get \( A=3\), \(B=2\), and \( C=−3\).
Thus,
\[ \dfrac{2x−3}{x^3+x}=\dfrac{3x+2}{x^2+1}−\dfrac{3}{x}.\nonumber \]
Substituting back into the integral, we obtain
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{2x−3}{x^3+x}\,dx & = & \displaystyle \int \left(\dfrac{3x+2}{x^2+1}−\dfrac{3}{x}\right)\,dx \\
& = & 3 \displaystyle \int \dfrac{x}{x^2+1}\,dx + 2 \displaystyle \int \dfrac{1}{x^2+1}\,dx−3 \displaystyle \int \dfrac{1}{x}\,dx \\
& = & \dfrac{3}{2}\ln ∣x^2+1∣+2 \tan^{−1}x−3\ln |x|+C. \\
\end{array} \nonumber \]
Note: We may rewrite \( \ln ∣x^2+1∣=\ln (x^2+1)\), if we wish to do so, since \( x^2+1 \gt 0\); however, this is not required.
Evaluate
\[\int \dfrac{dx}{x^3−8}. \nonumber \]
Solution
We can start by factoring \( x^3−8=(x−2)(x^2+2x+4)\). We see that the quadratic factor \( x^2+2x+4\) is irreducible since \( 2^2−4(1)(4) = −12 \lt 0\). Using the decomposition described in the problem-solving strategy, we get
\[ \dfrac{1}{(x−2)(x^2+2x+4)}=\dfrac{A}{x−2}+\dfrac{Bx+C}{x^2+2x+4}. \nonumber \]
Multiplying both sides by the LCD, we arrive at
\[ 1=A(x^2+2x+4)+(Bx+C)(x−2). \nonumber \]
Applying either method to find the constants, we get \( A=\frac{1}{12}\), \(B=−\frac{1}{12}\), and \( C=−\frac{1}{3}\).
Rewriting \( \displaystyle \int \frac{dx}{x^3−8}\), we have
\[ \int \dfrac{\,dx}{x^3−8}=\dfrac{1}{12}\int \dfrac{1}{x−2}\,dx−\dfrac{1}{12}\int \dfrac{x+4}{x^2+2x+4}\,dx. \nonumber \]
We can see that
\[ \int \dfrac{1}{x−2}\,dx=\ln |x−2|+C,\nonumber \]
but
\[ \int \dfrac{x+4}{x^2+2x+4}\,dx \nonumber \]
requires a bit more effort. Let’s begin by completing the square on \( x^2+2x+4\) to obtain
\[ x^2+2x+4=(x+1)^2+3. \nonumber \]
By letting \( u=x+1\) and consequently \( du=\,dx,\) we see that
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{x+4}{x^2+2x+4}\,dx & = & \displaystyle \int \dfrac{x+4}{(x+1)^2+3}\,dx \\
& = & \displaystyle \int \dfrac{u+3}{u^2+3}\,du \\
& = & \displaystyle \int \dfrac{u}{u^2+3}\,du + \displaystyle \int \dfrac{3}{u^2+3}\, du \\
& = & \dfrac{1}{2}\ln ∣u^2+3∣+\dfrac{3}{\sqrt{3}}\tan^{−1}\left(\dfrac{u}{\sqrt{3}}\right)+C \\
& = & \dfrac{1}{2}\ln ∣x^2+2x+4∣+\sqrt{3}\tan^{−1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C \\
\end{array} \nonumber \]
Substituting back into the original integral and simplifying gives
\[ \int \dfrac{dx}{x^3−8}=\dfrac{1}{12}\ln |x−2|−\dfrac{1}{24}\ln |x^2+2x+4|−\dfrac{\sqrt{3}}{12}\tan^{−1}\left(\dfrac{x+1}{\sqrt{3}}\right)+C. \nonumber \]
Here again, we can drop the absolute value if we wish to do so, since \( x^2+2x+4 \gt 0\) for all \( x\).
Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of \( f(x)=\frac{x^2}{(x^2+1)^2}\) and the \(x\)-axis over the interval \( [0,1]\) about the \(y\)-axis.
Solution
Let’s begin by sketching the region to be revolved (see Figure \(\PageIndex{1}\)). From the sketch, we see that the shell method is a good choice for solving this problem.
The volume is given by
\[ V=2 \pi \int ^1_0 x \cdot \dfrac{x^2}{(x^2+1)^2}\,dx = 2 \pi \int ^1_0 \dfrac{x^3}{(x^2+1)^2}\,dx. \nonumber \]
Since \( \deg((x^2+1)^2) = 4 \gt 3 = \deg(x^3)\), we can proceed with partial fraction decomposition. Note that \( (x^2+1)^2\) is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get
\[ \dfrac{x^3}{(x^2+1)^2} = \dfrac{Ax+B}{x^2+1}+\dfrac{Cx+D}{(x^2+1)^2}. \nonumber \]
Multiplying both sides by the LCD gives
\[ x^3=(Ax+B)(x^2+1)+Cx+D. \nonumber \]
Solving, we obtain \( A=1\), \(B=0\), \(C=−1\), and \( D=0\). Substituting back into the integral, we have
\[ V=2 \pi \int _0^1\dfrac{x^3}{(x^2+1)^2}\,dx=2 \pi \int _0^1\left(\dfrac{x}{x^2+1}−\dfrac{x}{(x^2+1)^2}\right)\,dx=2 \pi \left(\dfrac{1}{2}\ln (x^2+1)+\dfrac{1}{2} \cdot \dfrac{1}{x^2+1}\right)\bigg|^1_0= \pi \left(\ln 2−\tfrac{1}{2}\right). \nonumber \]
Set up the partial fraction decomposition for \[ \int \dfrac{x^2+3x+1}{(x+2)(x−3)^2(x^2+4)^2}\,dx. \nonumber \]
- Hint
-
Use the problem-solving strategy.
- Answer
-
\[ \dfrac{x^2+3x+1}{(x+2)(x−3)^2(x^2+4)^2}=\dfrac{A}{x+2}+\dfrac{B}{x−3}+\dfrac{C}{(x−3)^2}+\dfrac{Dx+E}{x^2+4}+\dfrac{Fx+G}{(x^2+4)^2} \nonumber \]
Rationalizing Substitutions
Before we leave our exploration of integration of rational functions via partial fraction decomposition, it's time we introduce one final integration technique. In truth, this is actually just a simple \( u \)-substitution, but it often leads to a rational function requiring partial fraction decomposition (hence, it appears at the end of this section) and the approach to computing the differential might be slightly different from what you are used to. This technique involves converting a non-rational function into a rational function via a specialized use of the Substitution Method. We call this method the Rationalizing Substitution Method. While there is no "hard and fast" rule to how or when to apply this method, it is a good technique to try if an integrand contains an expression of the form \( \sqrt[n]{g(x)} \).4 To illustrate the method, we will look at a single example (from two approaches).
Evaluate
\[ \int \dfrac{\sqrt{x}}{1 + \sqrt[4]{x}} \, dx \nonumber \]
Solution
I am going to approach this as if I were a student.
I begin by thinking of all the methods we have so far to help us evaluate integrals:
- Direct Integration: Since the integrand is not the derivative of an established, well-known function, we cannot directly integrate.
- Simple \( u \)-Substitution: If we let \( u = 1 + \sqrt[4]{x} \), then \( du = \frac{1}{4} x^{-3/4} \, dx \). While this might seem messy, with a little creativity (and adjusting how we view \( u \)-substitution), it is a viable tactic.
- Trigonometric Integrals: Since the integrand is currently not the product of powers of trigonometric functions, this technique is not viable.
- Trigonometric Substitution: As written, this does not look to have a sum or difference of squares, so it doesn't look to be a candidate for a trigonometric substitution.
- Partial Fraction Decomposition: The integrand is definitely not a rational function, so this method is out of the question.
So, it looks as though a simple \( u \)-substitution is our best bet; however, when we did the "scratch work," we arrived at \( du = \frac{1}{4} x^{-3/4} \, dx \). Is there a better substitution?
The simple answer to this question is, "No, but... there is a better way to write the \( u \)-substitution."
The following approach is incredibly useful, so pay close attention.
Let
\[ u = 1 + \sqrt[4]{x}. \label{2.4.9.a} \]
Before finding the differential, \( du \), let's get rid of radicals by isolating \( \sqrt[4]{x} \) in Equation \( \ref{2.4.9.a} \) and raising both sides of to the fourth power. Therefore,
\[ u - 1 = \sqrt[4]{x} \implies (u - 1)^4 = x. \label{2.4.9.b} \]
Then
\[ 4 (u - 1)^3 \, du = dx. \nonumber \]
Also, looking at the original integrand, we have \( \sqrt{x} \) in the numerator, so taking the square root of both sides of the result in Equation \( \ref{2.4.9.b} \), we get
\[ (u - 1)^2 = \sqrt{x}. \nonumber \]
Therefore,
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{\sqrt{x}}{1 + \sqrt[4]{x}} \, dx & = & \displaystyle \int \dfrac{(u - 1)^2}{u} \cdot 4 (u - 1)^3 \, du \\
& = & 4 \displaystyle \int \dfrac{(u - 1)^5}{u} \, du \\
& = & 4 \displaystyle \int \dfrac{u^5 - 5u^4 + 10 u^3 - 10u^2 + 5 u - 1}{u} \, du \\
& = & 4 \displaystyle \int u^4 - 5u^3 + 10 u^2 - 10u + 5 - \dfrac{1}{u} \, du \\
& = & 4 \left( \dfrac{1}{5}u^5 - \dfrac{5}{4} u^4 + \dfrac{10}{3} u^3 - 5 u^2 + 5u - \ln |u| \right) + C \\
& = & \dfrac{4}{5}u^5 - 5 u^4 + \dfrac{40}{3} u^3 - 20 u^2 + 20u - 4\ln |u| + C \\
& = & \dfrac{4}{5}\left( 1 + \sqrt[4]{x} \right)^5 - 5 \left( 1 + \sqrt[4]{x} \right)^4 + \dfrac{40}{3} \left( 1 + \sqrt[4]{x} \right)^3 - 20 \left( 1 + \sqrt[4]{x} \right)^2 + 20\left( 1 + \sqrt[4]{x} \right) - 4\ln \left( 1 + \sqrt[4]{x} \right) + C \\
\end{array} \nonumber \]
It's important to mention a couple important points about Example \( \PageIndex{9} \). First, the binomial expansion on the third line of the last set of computations is achieved quickly using either the Binomial Theorem (from Precalculus) or Pascal's Triangle (also from Precalculus). Knowing both of these concepts is going to be critical as we move forward in this course.
The second detail is that our choice of substitution was not unique. In fact, as is often the case with integration, we had several options.
For example, we could have chosen \( u = \sqrt[4]{x} \) so that \( u^2 = \sqrt{x} \) and \( u^4 = x \). This would yield \( 4 u^3 \, du = dx \). Substituting all of this into the original integral, we get
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{\sqrt{x}}{1 + \sqrt[4]{x}} \, dx & = & \displaystyle \int \dfrac{u^2}{1 + u} \cdot 4 u^3 \, du \\
& = & 4 \displaystyle \int \dfrac{u^5}{1 + u} \, du \\
\end{array} \nonumber \]
Using long division, we get
\[ \begin{array}{rcl}
\displaystyle \int \dfrac{\sqrt{x}}{1 + \sqrt[4]{x}} \, dx & = & 4 \displaystyle \int \dfrac{u^5}{1 + u} \, du \\
& = & 4 \displaystyle \int u^4 - u^3 + u^2 - u + 1 - \dfrac{1}{1 + u} \, du \\
& = & 4 \left( \dfrac{1}{5} u^5 - \dfrac{1}{4} u^4 + \dfrac{1}{3} u^3 - \dfrac{1}{2} u^2 + u - \ln|1 + u| \right) + C \\
& = & \dfrac{4}{5} x^{5/4} - x + \dfrac{4}{3} x^{3/4} - 2 x^{1/2} + 4 x^{1/4} - 4 \ln|1 + \sqrt[4]{x}| + C \\
\end{array} \nonumber \]
Notice that the answer in this last computation does not look like the answer we arrived at in Example \( \PageIndex{9} \). I assure you they are equivalent, but it's best you check by taking their derivatives and seeing if they match.
4 Of all of the methods we have covered so far, this method is the most heuristic. Try as many problems as possible that require (or can use) this technique to get a more natural understanding of when it is applicable.
Key Concepts
- Partial fraction decomposition is a technique used to break down a rational function into a sum of simple rational functions that can be integrated using previously learned techniques.
- When applying partial fraction decomposition, we must make sure that the degree of the numerator is less than the degree of the denominator. If not, we need to perform long division before attempting partial fraction decomposition.
- The form the decomposition takes depends on the type of factors in the denominator. The types of factors include nonrepeated linear factors, repeated linear factors, nonrepeated irreducible quadratic factors, and repeated irreducible quadratic factors.
Glossary
- partial fraction decomposition
- a technique used to break down a rational function into the sum of simple rational functions