{"id":716,"date":"2025-06-20T17:07:14","date_gmt":"2025-06-20T17:07:14","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=716"},"modified":"2025-07-17T15:28:20","modified_gmt":"2025-07-17T15:28:20","slug":"partial-fractions-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/partial-fractions-learn-it-4\/","title":{"raw":"Partial Fractions: Learn It 4","rendered":"Partial Fractions: Learn It 4"},"content":{"raw":"

Partial Fraction Decomposition<\/h2>\r\n

The General Method<\/h3>\r\nNow that we are beginning to get the idea of how the technique of partial fraction decomposition works, let\u2019s outline the basic method in the following problem-solving strategy.\r\n\r\n
Problem-Solving Strategy: Partial Fraction Decomposition<\/strong>\r\n
    \r\n \t
  1. 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.<\/li>\r\n \t
  2. 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.<\/li>\r\n \t
  3. 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].\r\n
      \r\n \t
    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\r\n<\/span>\r\n
      [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].<\/div><\/li>\r\n \t
    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\r\n<\/span>\r\n
      [latex]\\frac{{A}_{1}}{ax+b}+\\frac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\cdots +\\frac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex].<\/div><\/li>\r\n \t
    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\r\n<\/span>\r\n
      [latex]\\frac{Ax+B}{a{x}^{2}+bx+c}[\/latex].<\/div><\/li>\r\n \t
    4. For each repeated irreducible quadratic factor [latex]{\\left(a{x}^{2}+bx+c\\right)}^{n}[\/latex], the decomposition must include\r\n<\/span>\r\n
      [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].<\/div><\/li>\r\n \t
    5. After the appropriate decomposition is determined, solve for the constants.<\/li>\r\n \t
    6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/section>Now let\u2019s 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\u2014that is, if [latex]{b}^{2}-4ac<0[\/latex].\r\n\r\n
      \r\n
      \r\n

      Evaluate [latex]\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n

      \r\n

      Since [latex]\\text{deg}\\left(2x - 3\\right)<\\text{deg}\\left({x}^{3}+x\\right)[\/latex], factor the denominator and proceed with partial fraction decomposition. Since [latex]{x}^{3}+x=x\\left({x}^{2}+1\\right)[\/latex] contains the irreducible quadratic factor [latex]{x}^{2}+1[\/latex], include [latex]\\frac{Ax+B}{{x}^{2}+1}[\/latex] as part of the decomposition, along with [latex]\\frac{C}{x}[\/latex] for the linear term [latex]x[\/latex]. Thus, the decomposition has the form<\/p>\r\n\r\n

      [latex]\\frac{2x - 3}{x\\left({x}^{2}+1\\right)}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{C}{x}[\/latex].<\/div>\r\n \r\n

      After getting a common denominator and equating the numerators, we obtain the equation<\/p>\r\n\r\n

      [latex]2x - 3=\\left(Ax+B\\right)x+C\\left({x}^{2}+1\\right)[\/latex].<\/div>\r\n \r\n

      Solving for [latex]A,B[\/latex], and [latex]C[\/latex], we get [latex]A=3[\/latex], [latex]B=2[\/latex], and [latex]C=-3[\/latex].<\/p>\r\n

      Thus,<\/p>\r\n\r\n

      [latex]\\frac{2x - 3}{{x}^{3}+x}=\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}[\/latex].<\/div>\r\n \r\n

      Substituting back into the integral, we obtain<\/p>\r\n\r\n

      [latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx}& ={\\displaystyle\\int \\left(\\frac{3x+2}{{x}^{2}+1}-\\frac{3}{x}\\right)dx}\\hfill & & & \\\\ & =3{\\displaystyle\\int \\frac{x}{{x}^{2}+1}dx+2\\displaystyle\\int \\frac{1}{{x}^{2}+1}dx - 3\\displaystyle\\int \\frac{1}{x}dx}\\hfill & & & \\text{Split up the integral.}\\hfill \\\\ & =\\frac{3}{2}\\text{ln}|{x}^{2}+1|+2{\\tan}^{-1}x - 3\\text{ln}|x|+C.\\hfill & & & \\text{Evaluate each integral.}\\end{array}[\/latex]<\/div>\r\n \r\n

      Note<\/em>: We may rewrite [latex]\\text{ln}|{x}^{2}+1|=\\text{ln}\\left({x}^{2}+1\\right)[\/latex], if we wish to do so, since [latex]{x}^{2}+1>0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

      \r\n
      \r\n

      Evaluate [latex]\\displaystyle\\int \\frac{dx}{{x}^{3}-8}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n

      \r\n

      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<\/p>\r\n\r\n

      [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].<\/div>\r\n \r\n

      After obtaining a common denominator and equating the numerators, this becomes<\/p>\r\n\r\n

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

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

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

      [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].<\/div>\r\n \r\n

      We can see that<\/p>\r\n

      [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\u2019s begin by completing the square on [latex]{x}^{2}+2x+4[\/latex] to obtain<\/p>\r\n\r\n

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

      By letting [latex]u=x+1[\/latex] and consequently [latex]du=dx[\/latex], we see that<\/p>\r\n\r\n

      [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]<\/div>\r\n \r\n

      Substituting back into the original integral and simplifying gives<\/p>\r\n\r\n

      [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].<\/div>\r\n \r\n

      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].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

      \r\n
      \r\n

      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<\/em>-axis over the interval [latex]\\left[0,1\\right][\/latex] about the y<\/em>-axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n

      \r\n

      Let\u2019s begin by sketching the region to be revolved (see Figure 1). From the sketch, we see that the shell method is a good choice for solving this problem.<\/p>\r\n\r\n

      <\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"276\"]\"This Figure 1. We can use the shell method to find the volume of revolution obtained by revolving the region shown about the y-axis.[\/caption]<\/figure>\r\n

      The volume is given by<\/p>\r\n\r\n

      [latex]V=2\\pi {\\displaystyle\\int }_{0}^{1}x\\cdot \\frac{{x}^{2}}{{\\left({x}^{2}+1\\right)}^{2}}dx=2\\pi {\\displaystyle\\int }_{0}^{1}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx[\/latex].<\/div>\r\n \r\n

      Since [latex]\\text{deg}\\left({\\left({x}^{2}+1\\right)}^{2}\\right)=4>3=\\text{deg}\\left({x}^{3}\\right)[\/latex], we can proceed with partial fraction decomposition. Note that [latex]{\\left({x}^{2}+1\\right)}^{2}[\/latex] is a repeated irreducible quadratic. Using the decomposition described in the problem-solving strategy, we get<\/p>\r\n\r\n

      [latex]\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}=\\frac{Ax+B}{{x}^{2}+1}+\\frac{Cx+D}{{\\left({x}^{2}+1\\right)}^{2}}[\/latex].<\/div>\r\n \r\n

      Finding a common denominator and equating the numerators gives<\/p>\r\n\r\n

      [latex]{x}^{3}=\\left(Ax+B\\right)\\left({x}^{2}+1\\right)+Cx+D[\/latex].<\/div>\r\n \r\n

      Solving, we obtain [latex]A=1[\/latex], [latex]B=0[\/latex], [latex]C=-1[\/latex], and [latex]D=0[\/latex]. Substituting back into the integral, we have<\/p>\r\n\r\n

      [latex]\\begin{array}{cc}\\hfill V& =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\frac{{x}^{3}}{{\\left({x}^{2}+1\\right)}^{2}}dx\\hfill \\\\ & =2\\pi \\underset{0}{\\overset{1}{\\displaystyle\\int }}\\left(\\frac{x}{{x}^{2}+1}-\\frac{x}{{\\left({x}^{2}+1\\right)}^{2}}\\right)dx\\hfill \\\\ & =2\\pi \\left(\\frac{1}{2}\\text{ln}\\left({x}^{2}+1\\right)+\\frac{1}{2}\\cdot \\frac{1}{{x}^{2}+1}\\right)|{}_{\\begin{array}{c}\\\\ 0\\end{array}}^{\\begin{array}{c}1\\\\ \\end{array}}\\hfill \\\\ & =\\pi \\left(\\text{ln}2-\\frac{1}{2}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

      Partial Fraction Decomposition<\/h2>\n

      The General Method<\/h3>\n

      Now that we are beginning to get the idea of how the technique of partial fraction decomposition works, let\u2019s outline the basic method in the following problem-solving strategy.<\/p>\n

      Problem-Solving Strategy: Partial Fraction Decomposition<\/strong><\/p>\n
        \n
      1. 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.<\/li>\n
      2. 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.<\/li>\n
      3. 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].\n\n\n
          \n
        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
          \n<\/span><\/p>\n
          [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].<\/div>\n<\/li>\n
        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
          \n<\/span><\/p>\n
          [latex]\\frac{{A}_{1}}{ax+b}+\\frac{{A}_{2}}{{\\left(ax+b\\right)}^{2}}+\\cdots +\\frac{{A}_{n}}{{\\left(ax+b\\right)}^{n}}[\/latex].<\/div>\n<\/li>\n
        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
          \n<\/span><\/p>\n
          [latex]\\frac{Ax+B}{a{x}^{2}+bx+c}[\/latex].<\/div>\n<\/li>\n
        4. For each repeated irreducible quadratic factor [latex]{\\left(a{x}^{2}+bx+c\\right)}^{n}[\/latex], the decomposition must include
          \n<\/span><\/p>\n
          [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].<\/div>\n<\/li>\n
        5. After the appropriate decomposition is determined, solve for the constants.<\/li>\n
        6. Last, rewrite the integral in its decomposed form and evaluate it using previously developed techniques or integration formulas.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/section>\n

          Now let\u2019s 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\u2014that is, if [latex]{b}^{2}-4ac<0[\/latex].\n\n\n\n

          \n
          \n

          Evaluate [latex]\\displaystyle\\int \\frac{2x - 3}{{x}^{3}+x}dx[\/latex].<\/p>\n<\/div>\n