{"id":714,"date":"2025-06-20T17:07:08","date_gmt":"2025-06-20T17:07:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=714"},"modified":"2025-07-17T15:17:33","modified_gmt":"2025-07-17T15:17:33","slug":"partial-fractions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/partial-fractions-learn-it-2\/","title":{"raw":"Partial Fractions: Learn It 2","rendered":"Partial Fractions: Learn It 2"},"content":{"raw":"

Nonrepeated Linear Factors<\/h2>\r\n

When you're ready to use partial fraction decomposition, your first step is always the same: factor the denominator<\/strong> [latex]Q(x)[\/latex]. The simplest case occurs when [latex]Q(x)[\/latex] factors into distinct linear factors (no repeated factors).<\/p>\r\n\r\n

\r\n

distinct linear factors<\/h3>\r\n

If [latex]Q(x)[\/latex] factors as [latex]({a}_{1}x+{b}_{1})({a}_{2}x+{b}_{2})\\ldots({a}_{n}x+{b}_{n})[\/latex], where each factor appears only once, then:<\/p>\r\n

[latex]\\frac{P(x)}{Q(x)}=\\frac{A_1}{a_1x+b_1}+\\frac{A_2}{a_2x+b_2}+\\cdots +\\frac{A_n}{a_nx+b_n}[\/latex]<\/p>\r\nThe constants [latex]A_1, A_2, \\ldots, A_n[\/latex] are what you need to find.\r\n\r\n<\/section>Each distinct linear factor in the denominator gets its own fraction with a constant in the numerator.\r\n\r\n

If [latex]Q(x) = (x-1)(x+3)(2x-5)[\/latex], then your partial fraction setup would be:\r\n

[latex]\\frac{P(x)}{(x-1)(x+3)(2x-5)} = \\frac{A}{x-1} + \\frac{B}{x+3} + \\frac{C}{2x-5}[\/latex]<\/p>\r\n\r\n<\/section>

The mathematical proof that these constants [latex]A_1, A_2, \\ldots, A_n[\/latex] always exist is complex and beyond what you need for this course. The important thing is learning how to find them!<\/section>
\r\n

Evaluate [latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx[\/latex].<\/p>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n

\r\n

Since [latex]\\text{deg}\\left(3x+2\\right)<\\text{deg}\\left({x}^{3}-{x}^{2}-2x\\right)[\/latex], we begin by factoring the denominator of [latex]\\frac{3x+2}{{x}^{3}-{x}^{2}-2x}[\/latex]. We can see that [latex]{x}^{3}-{x}^{2}-2x=x\\left(x - 2\\right)\\left(x+1\\right)[\/latex]. Thus, there are constants [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] satisfying<\/p>\r\n\r\n

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

We must now find these constants. To do so, we begin by getting a common denominator on the right. Thus,<\/p>\r\n\r\n

[latex]\\frac{3x+2}{x\\left(x - 2\\right)\\left(x+1\\right)}=\\frac{A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)}{x\\left(x - 2\\right)\\left(x+1\\right)}[\/latex].<\/div>\r\n \r\n

Now, we set the numerators equal to each other, obtaining<\/p>\r\n\r\n

[latex]3x+2=A\\left(x - 2\\right)\\left(x+1\\right)+Bx\\left(x+1\\right)+Cx\\left(x - 2\\right)[\/latex].<\/div>\r\n \r\n

There are two different strategies for finding the coefficients [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex]. We refer to these as the method of equating coefficients<\/em><\/span> and the method of strategic substitution<\/em><\/span>.<\/p>\r\n\r\n

    \r\n \t
  • \r\n

    Rule: Method of Equating Coefficients<\/strong><\/span><\/p>\r\n

    Rewrite the previous equation in the form<\/p>\r\n\r\n

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

    Equating coefficients produces the system of equations<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill A+B+C& =\\hfill & 0\\hfill \\\\ \\hfill -A+B - 2C& =\\hfill & 3\\hfill \\\\ \\hfill -2A& =\\hfill & 2.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    To solve this system, we first observe that [latex]-2A=2\\Rightarrow A=-1[\/latex]. Substituting this value into the first two equations gives us the system<\/p>\r\n\r\n

    [latex]\\begin{array}{ccc}\\hfill B+C& =\\hfill & 1\\hfill \\\\ \\hfill B - 2C& =\\hfill & 2.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

    Multiplying the second equation by [latex]-1[\/latex] and adding the resulting equation to the first produces<\/p>\r\n\r\n

    [latex]-3C=1[\/latex],<\/div>\r\n \r\n

    which in turn implies that [latex]C=-\\frac{1}{3}[\/latex]. Substituting this value into the equation [latex]B+C=1[\/latex] yields [latex]B=\\frac{4}{3}[\/latex]. Thus, solving these equations yields [latex]A=-1[\/latex], [latex]B=\\frac{4}{3}[\/latex], and [latex]C=-\\frac{1}{3}[\/latex].<\/p>\r\n

    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.<\/p>\r\n<\/li>\r\n \t

  • \r\n

    Rule: Method of Strategic Substitution<\/strong><\/span><\/p>\r\n

    The method of strategic substitution is 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 [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] that satisfy the equation for all<\/em> values of [latex]x[\/latex]. That is, this equation must be true for any value of [latex]x[\/latex] we care to substitute into it. Therefore, by choosing values of [latex]x[\/latex] carefully and substituting them into the equation, we may find [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] easily. For example, if we substitute [latex]x=0[\/latex], the equation reduces to [latex]2=A\\left(-2\\right)\\left(1\\right)[\/latex]. Solving for [latex]A[\/latex] yields [latex]A=-1[\/latex]. Next, by substituting [latex]x=2[\/latex], the equation reduces to [latex]8=B\\left(2\\right)\\left(3\\right)[\/latex], or equivalently [latex]B=\\frac{4}{3}[\/latex]. Last, we substitute [latex]x=-1[\/latex] into the equation and obtain [latex]-1=C\\left(-1\\right)\\left(-3\\right)[\/latex]. Solving, we have [latex]C=-\\frac{1}{3}[\/latex].<\/p>\r\n

    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.<\/p>\r\n<\/li>\r\n<\/ul>\r\n

    Now that we have the values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], we rewrite the original integral:<\/p>\r\n\r\n

    [latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\displaystyle\\int \\left(\\text{-}\\frac{1}{x}+\\frac{4}{3}\\cdot \\frac{1}{\\left(x - 2\\right)}-\\frac{1}{3}\\cdot \\frac{1}{\\left(x+1\\right)}\\right)dx[\/latex].<\/div>\r\n \r\n

    Evaluating the integral gives us<\/p>\r\n\r\n

    [latex]\\displaystyle\\int \\frac{3x+2}{{x}^{3}-{x}^{2}-2x}dx=\\text{-}\\text{ln}|x|+\\frac{4}{3}\\text{ln}|x - 2|-\\frac{1}{3}\\text{ln}|x+1|+C[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>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.\r\n\r\n
    \r\n
    \r\n

    Evaluate [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n

    \r\n

    Since [latex]\\text{degree}\\left({x}^{2}+3x+1\\right)\\ge \\text{degree}\\left({x}^{2}-4\\right)[\/latex], we must perform long division of polynomials. This results in<\/p>\r\n\r\n

    [latex]\\frac{{x}^{2}+3x+1}{{x}^{2}-4}=1+\\frac{3x+5}{{x}^{2}-4}[\/latex].<\/div>\r\n \r\n

    Next, we perform partial fraction decomposition on [latex]\\frac{3x+5}{{x}^{2}-4}=\\frac{3x+5}{\\left(x+2\\right)\\left(x - 2\\right)}[\/latex]. We have<\/p>\r\n\r\n

    [latex]\\frac{3x+5}{\\left(x - 2\\right)\\left(x+2\\right)}=\\frac{A}{x - 2}+\\frac{B}{x+2}[\/latex].<\/div>\r\n \r\n

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

    [latex]3x+5=A\\left(x+2\\right)+B\\left(x - 2\\right)[\/latex].<\/div>\r\n \r\n

    Solving for [latex]A[\/latex] and [latex]B[\/latex] using either method, we obtain [latex]A=\\frac{11}{4}[\/latex] and [latex]B=\\frac{1}{4}[\/latex].<\/p>\r\n

    Rewriting the original integral, we have<\/p>\r\n\r\n

    [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=\\displaystyle\\int \\left(1+\\frac{11}{4}\\cdot \\frac{1}{x - 2}+\\frac{1}{4}\\cdot \\frac{1}{x+2}\\right)dx[\/latex].<\/div>\r\n \r\n

    Evaluating the integral produces<\/p>\r\n\r\n

    [latex]\\displaystyle\\int \\frac{{x}^{2}+3x+1}{{x}^{2}-4}dx=x+\\frac{11}{4}\\text{ln}|x - 2|+\\frac{1}{4}\\text{ln}|x+2|+C[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>As we see in the next example, it may be possible to apply the technique of partial fraction decomposition to a nonrational function. The trick is to convert the nonrational function to a rational function through a substitution.\r\n\r\n
    \r\n
    \r\n

    Evaluate [latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n

    \r\n

    Let\u2019s begin by letting [latex]u=\\sin{x}[\/latex]. Consequently, [latex]du=\\cos{x}dx[\/latex]. After making these substitutions, we have<\/p>\r\n\r\n

    [latex]\\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx=\\displaystyle\\int \\frac{du}{{u}^{2}-u}=\\displaystyle\\int \\frac{du}{u\\left(u - 1\\right)}[\/latex].<\/div>\r\n \r\n

    Applying partial fraction decomposition to [latex]\\frac{1}{u}\\left(u - 1\\right)[\/latex] gives [latex]\\frac{1}{u\\left(u - 1\\right)}=-\\frac{1}{u}+\\frac{1}{u - 1}[\/latex].<\/p>\r\n

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

    [latex]\\begin{array}{cc}\\hfill \\displaystyle\\int \\frac{\\cos{x}}{{\\sin}^{2}x-\\sin{x}}dx& =\\text{-}\\text{ln}|u|+\\text{ln}|u - 1|+C\\hfill \\\\ & =\\text{-}\\text{ln}|\\sin{x}|+\\text{ln}|\\sin{x} - 1|+C.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

    Nonrepeated Linear Factors<\/h2>\n

    When you’re ready to use partial fraction decomposition, your first step is always the same: factor the denominator<\/strong> [latex]Q(x)[\/latex]. The simplest case occurs when [latex]Q(x)[\/latex] factors into distinct linear factors (no repeated factors).<\/p>\n

    \n

    distinct linear factors<\/h3>\n

    If [latex]Q(x)[\/latex] factors as [latex]({a}_{1}x+{b}_{1})({a}_{2}x+{b}_{2})\\ldots({a}_{n}x+{b}_{n})[\/latex], where each factor appears only once, then:<\/p>\n

    [latex]\\frac{P(x)}{Q(x)}=\\frac{A_1}{a_1x+b_1}+\\frac{A_2}{a_2x+b_2}+\\cdots +\\frac{A_n}{a_nx+b_n}[\/latex]<\/p>\n

    The constants [latex]A_1, A_2, \\ldots, A_n[\/latex] are what you need to find.<\/p>\n<\/section>\n

    Each distinct linear factor in the denominator gets its own fraction with a constant in the numerator.<\/p>\n

    If [latex]Q(x) = (x-1)(x+3)(2x-5)[\/latex], then your partial fraction setup would be:<\/p>\n

    [latex]\\frac{P(x)}{(x-1)(x+3)(2x-5)} = \\frac{A}{x-1} + \\frac{B}{x+3} + \\frac{C}{2x-5}[\/latex]<\/p>\n<\/section>\n

    The mathematical proof that these constants [latex]A_1, A_2, \\ldots, A_n[\/latex] always exist is complex and beyond what you need for this course. The important thing is learning how to find them!<\/section>\n
    \n

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