6.4: Integration by Partial Fractions

Because the degrees of \((x - 2)\) and \((x + 3)\) are both one, they are linear factors. But the denominator's degree (two) is not greater than the numerator's degree (two), so we first apply polynomial division to get \[1 + \frac{8 - x}{(x - 2)(x + 3)} \pd\] Now we can split the remaining rational expression into a sum of partial fractions by writing \[\frac{8 - x}{(x - 2)(x + 3)} = \frac{A}{x - 2} + \frac{B}{x + 3} \pd \] Multiplying both sides by \((x - 2)(x + 3)\) clears the denominators, leaving \begin{equation} 8 - x = A(x + 3) + B(x - 2) \pd \label{eq:ex-1} \end{equation} There are two viable methods of solving for \(A\) and \(B.\)

Method 1 Expanding the right side of \(\eqref{eq:ex-1}\) gives \[ \ba 8 - x &= Ax + 3A + Bx - 2B \nl -x + 8 &= (A + B)x + (3A - 2B) \pd \ea \] Let's compare coefficients: Since \(-x = (A + B)x,\) it follows that \(-1 = A + B.\) Likewise, we have \(8 = 3A - 2B.\) The solution to the system of equations \[ \bc A + B = -1 \nl 3A - 2B = 8 \ec \] is \(A = 6/5\) and \(B = -11/5.\) Thus, \[1 + \frac{8 - x}{(x - 2)(x + 3)} = \boxed{1 + \frac{6/5}{x - 2} - \frac{11/5}{x + 3}}\]

Method 2 Because \(\eqrefer{eq:ex-1}\) is an identity, it's true for all \(x.\) To immediately solve for \(A\) and \(B,\) we can perform substitutions to equate one factor to \(0,\) leaving a single unknown. Letting \(x = 2\) causes the \(B\) to vanish, allowing us to solve for \(A\) easily: \[ \ba 8 - 2 &= A(2 + 3) + B(2 - 2) \nl 6 &= 5A + 0 \nl \implies A &= \tfrac{6}{5} \pd \ea \] Similarly, substituting \(x = -3\) gives \[ \ba 8 - (-3) &= A(-3 + 3) + B(-3 - 2) \nl 11 &= 0 - 5B \nl \implies B &= -\tfrac{11}{5} \pd \ea \] Thus, the partial fraction decomposition is \[\boxed{1 + \frac{6/5}{x - 2} - \frac{11/5}{x + 3}}\]

The denominator has a higher degree than the numerator, so we proceed with partial fraction decomposition. We factor the denominator as \(x(x + 2)(x + 5)\) and write \[ \frac{x^2 - 5x + 1}{x(x + 2)(x + 5)} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{x + 5} \pd \] Multiplying both sides by \(x(x + 2)(x + 5)\) clears the denominators: \[ x^2 - 5x + 1 = A(x + 2)(x + 5) + Bx(x + 5) + Cx(x + 2) \pd \] To solve for each coefficient, we let \(x\) be some value to zero out the other terms. For example, to solve for \(A\) we substitute \(x = 0\) to find \[ 1 = 10A + 0 + 0 \implies A = \tfrac{1}{10} \pd \] Similarly, to find \(B\) we let \(x = -2 \col\) \[15 = 0 - 6B + 0 \implies B = -\tfrac{5}{2} \pd\] Finally, substituting \(x = -5\) permits us to determine \(C \col\) \[51 = 0 + 0 + 15 C \implies C = \tfrac{17}{5} \pd\] Thus, \[ \ba \int \frac{x^2 - 5x + 1}{x^3 + 7x^2 + 10x} \di x &= \int \par{\frac{1/10}{x} + \frac{-5/2}{x + 2} + \frac{17/5}{x + 5}} \di x \nl &= \boxed{\tfrac{1}{10} \ln \abs x - \tfrac{5}{2} \ln \abs{x + 2} + \tfrac{17}{5} \ln \abs{x + 5} + c} \ea \]

Note that \(x^2 - a^2\) \(= (x + a)(x - a),\) so we write \[ \frac{1}{(x + a)(x - a)} = \frac{A}{x + a} + \frac{B}{x - a} \pd \] Multiplying both sides by \((x + a)(x - a)\) produces \[1 = A(x - a) + B(x + a) \pd\] Substituting \(x = a\) gives \[ 1 = 0 + B(2a) \implies B = \frac{1}{2a} \pd \] Likewise, letting \(x = -a\) yields \[1 = A(-2a) + 0 \implies A = -\frac{1}{2a} \pd\] Hence, \[ \ba \int \frac{1}{x^2 - a^2} \di x &= \int \par{\frac{-1/2a}{x + a} + \frac{1/2a}{x - a}} \di x \nl &= -\frac{1}{2a} \ln \abs{x + a} + \frac{1}{2a} \ln \abs{x - a} + c \nl &= \boxed{\frac{1}{2a} \ln \abs{\frac{x - a}{x + a}} + c} \ea \] The last step is true by the logarithm law \(\ln x - \ln y\) \(= \ln(x/y).\)

The denominator \((x^3 - 1)\) is a difference of cubes, which we factor as \[(x - 1)(x^2 + x + 1) \cma\] where \((x - 1)\) is a linear factor and \((x^2 + x + 1)\) is an irreducible quadratic factor. Therefore, the partial fraction decomposition takes the shape \[ \frac{4x + 2}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \pd \] Multiplying both sides by \((x - 1)(x^2 + x + 1)\) gives \[ 4x + 2 = A(x^2 + x + 1) + Bx(x - 1) + C(x - 1) \pd \] Next we solve for \(A,\) \(B,\) and \(C\) by performing substitutions. Letting \(x = 1\) eliminates terms involving \(B\) and \(C,\) yielding \[6 = 3A + 0 + 0 \implies A = 2 \pd\] Moreover, letting \(x = 0\) gives \[ \ba 2 &= A + 0 - C \nl 2 &= 2 - C \nl \implies C &= 0 \pd \ea \] We now have \[4x + 2 = 2(x^2 + x + 1) + Bx(x - 1) \pd\] We can't zero out \((x^2 + x + 1)\) to solve for \(B.\) Instead, we just let \(x\) be any value—say, \(x = 2 \col\) \[10 = 14 + 2B \implies B = -2 \pd\] Thus, our partial fraction decomposition is \[ \frac{4x + 2}{x^3 - 1} = \boxed{\frac{2}{x - 1} - \frac{2x}{x^2 + x + 1}} \]

The factor \((x + 4)\) is linear, while \(x^3\) is repeated and \((2x^2 + 4x + 7)^2\) is a repeated irreducible quadratic factor. Hence, the partial fraction decomposition is of the form \[\boxed{\frac{A}{x + 4} + \frac{B}{x} + \frac{C}{x^2} + \frac{D}{x^3} + \frac{Ex + F}{2x^2 + 4x + 7} + \frac{Gx + H}{\par{2x^2 + 4x + 7}^2}}\]

The factor \((x + 2)^2\) is repeated, so we write \[ \frac{1}{x(x + 2)^2} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2} \pd \] Multiplying each term by \(x(x + 2)^2\) clears the denominators, giving \[ 1 = A(x + 2)^2 + Bx(x + 2) + Cx \pd \] We now solve for \(A,\) \(B,\) and \(C\) by using substitutions to zero out terms. Letting \(x = 0,\) we get \[1 = 4A + 0 + 0 \implies A = \tfrac{1}{4} \pd\] Another convenient substitution is \(x = -2,\) which gives \[1 = 0 + 0 - 2 C \implies C = -\tfrac{1}{2} \pd\] Hence, we have \[1 = \tfrac{1}{4} (x + 2)^2 + Bx(x + 2) - \tfrac{1}{2} x \pd\] The remaining unknown is \(B,\) whose value we can't immediately attain by zeroing out the other terms. Instead, we just let \(x\) be any other value—say, \(x = -1 \col\) \[1 = \tfrac{1}{4} - B + \tfrac{1}{2} \implies B = -\tfrac{1}{4} \pd\] Thus, \[ \ba \int \frac{1}{x(x + 2)^2} \di x &= \int \frac{1/4}{x} + \frac{-1/4}{x + 2} + \frac{-1/2}{(x + 2)^2} \di x \nl &= \boxed{\tfrac{1}{4} \ln \abs{x} - \tfrac{1}{4} \ln \abs{x + 2} + \frac{1}{2x + 4} + c} \ea \]

REMARK The same result can be attained by treating \((x + 2)^2\) as an irreducible quadratic term and using the shape \[ \frac{1}{x(x + 2)^2} = \frac{A}{x} + \frac{Bx + C}{(x + 2)^2} \pd \]

The denominator contains two linear factors, \((x - 2)\) and \((2x - 6),\) and the irreducible quadratic factor \((x^2 + 16).\)

Partial Fraction Decomposition The partial fraction decomposition takes the following shape: \[ \frac{7 - 3x^2}{(x - 2)(2x - 6)(x^2 + 16)} = \frac{A}{x - 2} + \frac{B}{2x - 6} + \frac{Cx + D}{x^2 + 16} \pd \] Multiplying both sides by \((x - 2)(2x - 6)(x^2 + 16)\) yields \[ \small 7 - 3x^2 = A(2x - 6)(x^2 + 16) + B(x - 2)(x^2 + 16) + Cx(x - 2)(2x - 6) + D(x - 2)(2x - 6) \pd \] Let's perform substitutions to zero out some terms. For example, letting \(x = 2\) leaves only one term, producing \[-5 = -40A + 0 + 0 + 0 \implies A = \tfrac{1}{8} \pd\] Similarly, all but one term include the factor \((2x - 6),\) so substituting \(x = 3\) zeroes out most terms, leaving \[ -20 = 0 + 25B + 0 + 0 \implies B = -\tfrac{4}{5} \pd \] Now we have \[ \small 7 - 3x^2 = \tfrac{1}{8}(2x - 6)(x^2 + 16) - \tfrac{4}{5} (x - 2)(x^2 + 16) + Cx(x - 2)(2x - 6) + D(x - 2)(2x - 6) \pd \] To solve for \(C\) and \(D,\) we pick arbitrary values for \(x;\) for example, letting \(x = 0\) gives \[ 7 = \tfrac{1}{8}(-6)(16) - \tfrac{4}{5}(-2)(16) + 0 + 12D \implies D = -\tfrac{11}{20} \pd \] The equation therefore becomes \[ \small 7 - 3x^2 = \tfrac{1}{8}(2x - 6)(x^2 + 16) - \tfrac{4}{5} (x - 2)(x^2 + 16) + Cx(x - 2)(2x - 6) - \tfrac{11}{20} (x - 2)(2x - 6) \pd \] We now assign \(x\) any value to immediately solve for \(C\)—say, \(x = 1 \col\) \[ 4 = \tfrac{1}{8}(-4)(17) - \tfrac{4}{5}(-1)(17) + 4C - \tfrac{11}{20} (-1)(-4) \implies C = \tfrac{11}{40} \pd \]

Integration We have shown \[ \int \frac{7 - 3x^2}{(x - 2)(2x - 6)(x^2 + 16)} \di x = \int \frac{\frac{1}{8}}{x - 2} - \frac{\frac{4}{5}}{2x - 6} + \frac{\frac{11}{40} x - \frac{11}{20} }{x^2 + 16} \di x \pd \] Note that \[ \int \frac{\frac{11}{40}x - \frac{11}{20}}{x^2 + 16} \di x = \int \frac{\frac{11}{40}x}{x^2 + 16} \di x - \int \frac{\frac{11}{20}}{x^2 + 16} \di x \pd \] To evaluate the first integral on the right, we substitute \(u = x^2 + 16.\) Then \(\dd u = 2x \di x\) and so the integral becomes \[ \ba \int \frac{\tfrac{11}{40}}{u} \par{\frac{1}{2}} \di u &= \tfrac{11}{80} \ln \abs u \nl &= \tfrac{11}{80} \ln \par{x^2 + 16} \cma \ea \] where we neglected the constant of integration and dropped the absolute value bars because \((x^2 + 16)\) is always positive. To evaluate \(\int \frac{\frac{11}{20}}{x^2 + 16} \di x,\) we write \[\int \frac{\frac{11}{20}}{16 \parbr{\par{x/4}^2 + 1}} \di x = \int \frac{\frac{11}{320}}{\par{x/4}^2 + 1} \di x \pd\] Substituting \(u = x/4,\) we attain \(\dd u = \dd x /4\) and so the integral becomes \[ \ba \int \frac{\tfrac{11}{320} }{u^2 + 1} (4) \di u &= \tfrac{11}{80} \atan u \nl &= \tfrac{11}{80} \atan \par{\frac{x}{4}} \pd \ea \] (We have neglected the constant of integration.) Hence, \[ \small \ba \int \frac{7 - 3x^2}{(x - 2)(2x - 6)(x^2 + 16)} \di x &= \int \frac{\frac{1}{8}}{x - 2} - \frac{\frac{4}{5}}{2x - 6} + \frac{\frac{11}{40} x - \frac{11}{20} }{x^2 + 16} \di x \nl &= \boxed{\small \tfrac{1}{8} \ln \abs{x - 2} - \tfrac{2}{5} \ln \abs{2x - 6} + \tfrac{11}{80} \ln \par{x^2 + 16} - \tfrac{11}{80} \atan \par{\frac{x}{4}} + c} \ea \]