Partial Fractions: Learn It 4

Giselle Miller

Partial Fractions: Learn It 4

Partial Fraction Decomposition

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.

Problem-Solving Strategy: Partial Fraction Decomposition

Make sure that [latex]\text{degree}\left(P\left(x\right)\right)<\text{degree}\left(Q\left(x\right)\right)[/latex]. If not, perform long division of polynomials.
Factor [latex]Q\left(x\right)[/latex] into the product of linear and irreducible quadratic factors. An irreducible quadratic is a quadratic that has no real zeros.
Assuming that [latex]\text{deg}\left(P\left(x\right)\right)<\text{deg}\left(Q\left(x\right)\right)[/latex], the factors of [latex]Q\left(x\right)[/latex] determine the form of the decomposition of [latex]\frac{P\left(x\right)}{Q\left(x\right)}[/latex].
1. If [latex]Q\left(x\right)[/latex] can be factored as [latex]\left({a}_{1}x+{b}_{1}\right)\left({a}_{2}x+{b}_{2}\right)\ldots\left({a}_{n}x+{b}_{n}\right)[/latex], where each linear factor is distinct, then it is possible to find constants [latex]{A}_{1},{A}_{2},...{A}_{n}[/latex] satisfying
  
  [latex]\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{{a}_{1}x+{b}_{1}}+\frac{{A}_{2}}{{a}_{2}x+{b}_{2}}+\cdots +\frac{{A}_{n}}{{a}_{n}x+{b}_{n}}[/latex].
2. If [latex]Q\left(x\right)[/latex] contains the repeated linear factor [latex]{\left(ax+b\right)}^{n}[/latex], then the decomposition must contain
  
  [latex]\frac{{A}_{1}}{ax+b}+\frac{{A}_{2}}{{\left(ax+b\right)}^{2}}+\cdots +\frac{{A}_{n}}{{\left(ax+b\right)}^{n}}[/latex].
3. For each irreducible quadratic factor [latex]a{x}^{2}+bx+c[/latex] that [latex]Q\left(x\right)[/latex] contains, the decomposition must include
  
  [latex]\frac{Ax+B}{a{x}^{2}+bx+c}[/latex].
4. For each repeated irreducible quadratic factor [latex]{\left(a{x}^{2}+bx+c\right)}^{n}[/latex], the decomposition must include
  
  [latex]\frac{{A}_{1}x+{B}_{1}}{a{x}^{2}+bx+c}+\frac{{A}_{2}x+{B}_{2}}{{\left(a{x}^{2}+bx+c\right)}^{2}}+\cdots +\frac{{A}_{n}x+{B}_{n}}{{\left(a{x}^{2}+bx+c\right)}^{n}}[/latex].
5. After the appropriate decomposition is determined, solve for the constants.
6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.

Now let’s look at integrating a rational expression in which the denominator contains an irreducible quadratic factor. Recall that the quadratic [latex]a{x}^{2}+bx+c[/latex] is irreducible if [latex]a{x}^{2}+bx+c=0[/latex] has no real zeros—that is, if [latex]{b}^{2}-4ac<0[/latex].

Evaluate [latex]\displaystyle\int \frac{2x - 3}{{x}^{3}+x}dx[/latex].

Evaluate [latex]\displaystyle\int \frac{dx}{{x}^{3}-8}[/latex].

We can start by factoring [latex]{x}^{3}-8=\left(x - 2\right)\left({x}^{2}+2x+4\right)[/latex]. We see that the quadratic factor [latex]{x}^{2}+2x+4[/latex] is irreducible since [latex]{2}^{2}-4\left(1\right)\left(4\right)=-12<0[/latex]. Using the decomposition described in the problem-solving strategy, we get

[latex]\frac{1}{\left(x - 2\right)\left({x}^{2}+2x+4\right)}=\frac{A}{x - 2}+\frac{Bx+C}{{x}^{2}+2x+4}[/latex].

After obtaining a common denominator and equating the numerators, this becomes

[latex]1=A\left({x}^{2}+2x+4\right)+\left(Bx+C\right)\left(x - 2\right)[/latex].

Applying either method, we get [latex]A=\frac{1}{12},B=-\frac{1}{12},\text{and}C=-\frac{1}{3}[/latex].

Rewriting [latex]\displaystyle\int \frac{dx}{{x}^{3}-8}[/latex], we have

[latex]\displaystyle\int \frac{dx}{{x}^{3}-8}=\frac{1}{12}\displaystyle\int \frac{1}{x - 2}dx-\frac{1}{12}\displaystyle\int \frac{x+4}{{x}^{2}+2x+4}dx[/latex].

We can see that

[latex]\displaystyle\int \frac{1}{x - 2}dx=\text{ln}|x - 2|+C[/latex], but [latex]\displaystyle\int \frac{x+4}{{x}^{2}+2x+4}dx[/latex] requires a bit more effort. Let’s begin by completing the square on [latex]{x}^{2}+2x+4[/latex] to obtain

[latex]{x}^{2}+2x+4={\left(x+1\right)}^{2}+3[/latex].

By letting [latex]u=x+1[/latex] and consequently [latex]du=dx[/latex], we see that

[latex]\begin{array}{ccccc}\hfill {\displaystyle\int \frac{x+4}{{x}^{2}+2x+4}dx}& ={\displaystyle\int \frac{x+4}{{\left(x+1\right)}^{2}+3}dx}\hfill & & & \begin{array}{c}\text{Complete the square on the}\hfill \\ \text{denominator.}\hfill \end{array}\hfill \\ & ={\displaystyle\int \frac{u+3}{{u}^{2}+3}du}\hfill & & & \begin{array}{c}\text{Substitute}u=x+1,x=u - 1,\hfill \\ \text{and}du=dx.\hfill \end{array}\hfill \\ & ={\displaystyle\int \frac{u}{{u}^{2}+3}du+\displaystyle\int \frac{3}{{u}^{2}+3}du}\hfill & & & \text{Split the numerator apart.}\hfill \\ & =\frac{1}{2}\text{ln}|{u}^{2}+3|+\frac{3}{\sqrt{3}}{\tan}^{-1}\frac{u}{\sqrt{3}}+C\hfill & & & \text{Evaluate each integral.}\hfill \\ & =\frac{1}{2}\text{ln}|{x}^{2}+2x+4|+\sqrt{3}{\tan}^{-1}\left(\frac{x+1}{\sqrt{3}}\right)+C.\hfill & & & \begin{array}{c}\text{Rewrite in terms of}x\text{and}\hfill \\ \text{simplify.}\hfill \end{array}\hfill \end{array}[/latex]

Substituting back into the original integral and simplifying gives

[latex]{\displaystyle\int}\frac{dx}{{x}^{3}-8}=\frac{1}{12}\text{ln}|x - 2|-\frac{1}{24}\text{ln}|{x}^{2}+2x+4|-\frac{\sqrt{3}}{12}{\tan}^{-1}\left(\frac{x+1}{\sqrt{3}}\right)+C[/latex].

Here again, we can drop the absolute value if we wish to do so, since [latex]{x}^{2}+2x+4>0[/latex] for all [latex]x[/latex].

Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph of [latex]f\left(x\right)=\frac{{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}[/latex] and the x-axis over the interval [latex]\left[0,1\right][/latex] about the y-axis.