Partial fractions

Introduction to partial fractions

  • In this section, we show how to integrate a rational function \[ f(x)=\frac{p(x)}{q(x)}, \] provided that we can factorize the polynomial in the denominator \(q\) to a product of linear polynomials and irreducible quadratic polynomials.
    • To illustrate the method, observe that we have \[ \frac{1}{x-3}-\frac{2}{x+2}=\frac{x+2-2(x-3)}{(x-3)(x+2)}=\frac{-x+6}{x^2-x-6}. \]
    • Therefore, reversing the process, we get \[ \int\frac{-x+6}{x^2-x-6}\,\mathrm dx=\int\frac{1}{x-3}-\frac{2}{x+2}\,\mathrm dx=\ln|x-3|-2\ln|x+2|+C. \]
  • The method of Partial fractions works in case the degree of \(p\) is less than the degree of \(q\).
    • Recall that if we have \[ p(x)=a_nx^n+a_{n-1}x^{n-1}+\dotsb+a_1x+a_0 \] with \(a_n\ne0\), then we say that the degree of \(p(x)\) is \(n\), and we write \(\mathrm{deg}(p)=n\).
    • If we have \(\mathrm{deg}(p)<\mathrm{deg}(q)\), then we call the rational function \(f=\frac{p}{q}\) proper.
    • If \(f\) is not proper, that is if \(\mathrm{deg}(p)\ge\mathrm{deg}(q)\), then we call \(f\) improper.
    • If \(f\) is improper, then we have to take the preliminary step of performing long division of \(p\) by \(q\).

Introduction to partial fractions

  • For example, consider \(f(x)=\frac{x^3-2x}{x+1}\).
    • We perform long division: \[\begin{align*} &\hphantom{x+1)}\underline{x^2-x-1}\\ &x+1)x^3\hphantom{{}-x}-2x\\ &\hphantom{x+1)}\underline{x^3+x^2}\\ &\hphantom{x+1)x^3}-x^2-2x\\ &\hphantom{x+1)x^3}\underline{{}-x^2-\hphantom{2}x}\\ &\hphantom{x+1)x^3-x^2}-\hphantom{2}x\\ &\hphantom{x+1)x^3-x^2}\underline{{}-\hphantom{2}x-1}\\ &\hphantom{x+1)x^3-x^2-2x-{}}1 \end{align*}\]
    • This gives \(x^3-2x=(x+1)(x^2-x-1)+1\), and thus \(\frac{x^3-2x}{x+1}=x^2-x-1+\frac{1}{x+1}\).
    • Therefore, we get \[ \int\frac{x^3-2x}{x+1}\,\mathrm dx=\int x^2-x-1+\frac{1}{x+1}\,\mathrm dx=\frac13x^3-\frac12x^2-x+\ln|x+1|+C. \]

Case I. The denominator \(q\) is a product of distinct linear factors

  • Suppose that we have \[ q(x)=(a_1x+b_1)(a_2x+b_2)\dotsb(a_kx+b_k), \] where no factor is a constant multiple of another.
    • This means that if \(\frac{b_i}{a_i}=\frac{b_j}{a_j}\) for some \(1\le i,j\le k\), then we have \(i=j\).
    • In this case, there exist constants \(A_1,A_2,\dotsc,A_k\) such that \[ \frac{p(x)}{q(x)}=\frac{A_1}{a_1x+b_1}+\frac{A_2}{a_2x+b_2}+\dotsb+\frac{A_k}{a_kx+b_k}. \]
    • The method of Partial fractions amounts to finding the constants \(A_1,A_2,\dotsc,A_n\).
    • This can be done by i) multiplying the equation by \(q(x)\), ii) equating the coefficients in the polynomial equation we get, and iii) solving the linear equation we've gotten.

Case I. The denominator \(q\) is a product of distinct linear factors

  • \(\int\frac{x^2-5x+4}{x^3+5x^2+6x}\)
    • First, we need to factorize the denominator: \[ x^3+5x^2+6x=x(x^2+5x+6)=x(x+2)(x+3). \]
    • Then we need to multiply the equation \[ \frac{x^2-5x+4}{x^3+5x^2+6x}=\frac{A_1}{x}+\frac{A_2}{x+2}+\frac{A_3}{x+3} \] by \(q\).
    • We get \[\begin{align*} x^2-5x+4=&A_1(x+2)(x+3)+A_2x(x+3)+A_3x(x+2)\\ =&A_1(x^2+5x+6)+A_2(x^2+3x)+A_3(x^2+2x)\\ =&x^2(A_1+A_2+A_3)+x(5A_1+3A_2)+6A_1. \end{align*}\]

Case I. The denominator \(q\) is a product of distinct linear factors

  • Equating the coefficients, we get the system of linear equations \[\begin{align*} \hphantom{6}A_1+\hphantom{3}A_2+\hphantom{2}A_3&=1\\ 5A_1+3A_2+2A_3&=-5\\ 6A_1\hphantom{{}+3A_2+2A_3}&=4. \end{align*}\]
    • We immediately get \(A_1=\frac23\). Substituting this to the first two equations, we get \[\begin{align} \hphantom3A_2+\hphantom2A_3&=\frac13\tag{I}\\ 3A_2+2A_3&=-\frac{25}{3}\tag{II}. \end{align}\]
    • Using this, we get \[\begin{align*} II-2I:&A_2=-9\\ -II+3I:&A_3=\frac{28}{3}. \end{align*}\]
    • We have gotten \[ \frac{x^2-5x+4}{x^3+5x^2+6x}=\frac{A_1}{x}+\frac{A_2}{x+2}+\frac{A_3}{x+3}=\frac23\frac{1}{x}-9\frac{1}{x+2}+\frac{28}{3}\frac{1}{x+3}. \]
    • Now we can integrate: \[ \int\frac{x^2-5x+4}{x^3+5x^2+6x}\,\mathrm dx=\int\frac23\frac{1}{x}-9\frac{1}{x+2}+\frac{28}{3}\frac{1}{x+3}\,\mathrm dx=\frac23\ln|x|-9\ln|x+2|+\frac{28}{3}\ln|x+3|+C. \]
  • Exercises. 7.4: 7, 11, 15, 27

Case II. \(q\) is a product of linear factors, some of which are repeated

  • Suppose that the factor \((a_1x+b_1)\) is repeated \(r\) times; that is, \((a_1x+b_1)^r\) occurs in the factorization of \(q(x)\). Then we need to replace the single term \(\frac{A_1}{a_1x+b_1}\) by the sum \[ \frac{A_1}{a_1x+b_1}+\frac{A_2}{(a_1x+b_1)^2}+\dotsb+\frac{A_r}{(a_1x+b_1)^r}. \]
  • \(\int\frac{x^2+2x-3}{x^3-x^2-x+1}\,\mathrm dx\).
    • Since we have \(q(1)=0\), we know that \(x-1\) divides \(q(x)\).
    • Performing the division, we get \[ q(x)=(x-1)(x^2-1)=(x-1)^2(x+1). \]
    • Therefore, we need \[ \frac{x^2+2x-3}{(x-1)^2(x+1)}=\frac{A_1}{x-1}+\frac{A_2}{(x-1)^2}+\frac{A_3}{x+1}. \]
    • Multiplying the equation with \(q\), we get \[\begin{align*} x^2+2x-3=&A_1(x-1)(x+1)+A_2(x+1)+A_3(x-1)^2\\ =&A_1(x^2-1)+A_2(x+1)+A_3(x^2-2x+1)\\ =&x^2(A_1+A_3)+x(A_2-2A_3)+(-A_1+A_2+A_3). \end{align*}\]

Case II. \(q\) is a product of linear factors, some of which are repeated

  • This gives the system of linear equations \[\begin{align} \hphantom{-}A_1\hphantom{{}+A_2}+\hphantom{2}A_3&=1\tag{I}\\ \hphantom{-A_1+{}}A_2-2A_3&=2\tag{II}\\ -A_1+A_2+\hphantom{2}A_3&=-3.\tag{III} \end{align}\]
    • \(\mathrm{I}+\mathrm{III}\) gives an additional equation \[\begin{equation} A_2+2A_3=-2.\tag{IV} \end{equation}\]
    • Then \(\mathrm{II}+\mathrm{IV}\) gives \(A_2=0\), using which in II we get \(A_3=-1\), and substituting that into I gives \(A_1=2\).
    • Now we can integrate: \[ \int\frac{x^2+2x-3}{x^3-x^2-x+1}\,\mathrm dx=\int\frac{2}{x-1}-\frac{1}{x+1}\,\mathrm dx=2\ln|x-1|-\ln|x+1|+C. \]
  • Exercises. 7.4: 16, 21.

Case III. \(q\) contains irreducible quadratic factors, none of which is a constant multiple of the other

  • Consider the polynomial \(r(x)=ax^2+bx+c\).
    • Recall that if the discriminant is negative: \(b^2-4ac<0\), then \(r(x)\) has no real roots. In this case, we say that \(r\) is irreducible.
  • For every irreducible quadratic factor \(ax^2+bx+c\) of \(q(x)\), we need to include a term of the form \[ \frac{Ax+B}{ax^2+bx+c} \] in the Partial fractions formula.
    • After we've found the constants \(A\) and \(B\), we need to complete the square in the denominator, and use \[ \int\frac{Cx+D}{x^2+\alpha^2}\,\mathrm dx=\frac{C}{2}\ln|x^2+\alpha^2|+\frac{D}{\alpha}\tan^{-1}\frac{x}{\alpha}+K. \]

Case III. \(q\) contains irreducible quadratic factors, none of which is a constant multiple of the other

  • \(\int\frac{x^2-2x+3}{x^3+x^2+x}\).
    • We have \(q(x)=x(x^2+x+1)\). Since its discriminant is \(1-4<0\), the polynomial \(x^2+x+1\) is irreducible.
    • Therefore, we need to solve \[ \frac{x^2-2x+3}{x(x^2+x+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+x+1}. \]
    • Multiplying with \(q\), we get \[\begin{align*} x^2-2x+3=&A(x^2+x+1)+(Bx+C)x\\ =&x^2(A+B)+x(A+C)+A. \end{align*}\]
    • This gives the system of linear equations \[\begin{align*} A+B\hphantom{{}+C}&=3\\ A\hphantom{{}+B}+C&=-2\\ A\hphantom{{}+B+C}&=3, \end{align*}\] which gives \(A=3,\,B=0,\,C=-5\).

Case III. \(q\) contains irreducible quadratic factors, none of which is a constant multiple of the other

  • Now we can proceed: \[ \int\frac{x^2-2x+3}{x^3+x^2+x}\,\mathrm dx=\int\frac{3}{x}-\frac{5}{x^2+x+1}\,\mathrm dx \]
    • Let's integrate the second term separately. We have \(x^2+x+1=(x+\frac12)^2+\frac34\). Therefore, the substitution \(u=x+\frac12\) gives \[ \int\frac{1}{x^2+x+1}\,\mathrm dx=\frac{1}{u^2+(\sqrt3/2)^2}\,\mathrm du=\frac{2}{\sqrt3}\tan^{-1}\frac{2u}{\sqrt3}+K=\frac{2}{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}+K. \]
    • This gives the final answer: \[ \int\frac{3}{x}-\frac{5}{x^2+x+1}\,\mathrm dx=3\ln|x|-\frac{10}{\sqrt3}\tan^{-1}\frac{2x+1}{\sqrt3}+K. \]
  • Exercises. 7.4: 23, 25, 32.

Case IV. \(q\) contains a repeated irreducible quadratic factor

  • Suppose that \(q(x)\) has the factor \((ax^2+bx+c)^r\), where \(b^2-4ac<0\). Then in the Partial fractions formula, the term \[ \frac{Ax+B}{ax^2+bx+c} \] needs to be replaced by the sum \[ \frac{A_1x+B_1}{ax^2+bx+c}+\frac{A_2x+B_2}{(ax^2+bx+c)^2}+\dotsb+\frac{A_rx+B_r}{(ax^2+bx+c)^r}. \]
    • To integrate the term \(\int\frac{1}{(ax^2+bx+c)^2}\,\mathrm dx\), you need to first complete the square to get \(\int\frac{1}{(x^2+\alpha^2)^r}\,\mathrm dx\), and then substitute \(x=\alpha\tan\theta\).

Case IV. \(q\) contains a repeated irreducible quadratic factor

  • \(\int\frac{x+3}{(2x^2+x+2)^2}\,\mathrm dx\).
    • We need to solve \[ \frac{x+3}{(2x^2+x+2)^2}=\frac{A_1x+B_1}{2x^2+x+2}+\frac{A_2x+B_2}{(2x^2+x+2)^2}. \]
    • Multiplying with \(q\), we get \[\begin{align*} x+3=&(A_1x+B_1)(2x^2+x+2)+A_2x+B_2\\ =&x^32A_1+x^2(A_1+2B_1)+x(2A_1+2B_1+A_2)+2B_1+B_2. \end{align*}\]
    • Equating the coefficients, we get \(A_1=0\), \(B_1=0\), \(A_2=1\), \(B_2=3\).
    • Therefore, we have \[ \int\frac{x+3}{(2x^2+x+2)^2}\,\mathrm dx=\int\frac{x+3}{(2x^2+x+2)^2}\,\mathrm dx. \]
    • Let's integrate the terms separately. First, we have \[ \int\frac{x}{(2x^2+x+2)^2}\,\mathrm dx\stackrel{u=2x^2+x+2}{=}\frac14\int u^{-2}\,\mathrm du=-\frac{1}{4(2x^2+x+2)}+C \]

Case IV. \(q\) contains a repeated irreducible quadratic factor

  • Now let's consider \(\int\frac{1}{(2x^2+x+2)^2}\,\mathrm dx\). First, to complete the square: \[ 2x^2+x+2=(\sqrt2x+2^{-3/2})^2+\frac{15}{8}. \]
    • Let \(\alpha=\sqrt\frac{15}{8}\), and take \[ \int\frac{1}{(2x^2+x+2)^2}\,\mathrm dx\stackrel{u=\sqrt2x+2^{-3/2}}{=}2^{-1/2}\int\frac{1}{(u^2+\alpha^2)^2}\,\mathrm du. \]
    • We can keep going by taking \(u=\alpha\theta\): \[ =2^{-1/2}\int\frac{\alpha^2(\sec\theta)^2}{\alpha^4(\sec\theta)^4}\,\mathrm d\theta=2^{-1/2}\alpha^{-2}\int(\cos\theta)^2\,\mathrm d\theta=2^{-3/2}\alpha^{-2}\int1+\cos2\theta\,\mathrm d\theta=2^{-3/2}\alpha^{-2}\left(\theta+\frac12\sin2\theta\right)+C. \]
    • Since we have \(\tan\theta=\frac{u}{\alpha}\), by drawing a right triangle with leg across \(\theta\) of length \(x\) and other leg of length \(\alpha\) to see that \[ \sin\theta=\frac{u}{\sqrt{u^2+\alpha^2}}\text{ and }\cos\theta=\frac{\alpha}{\sqrt{u^2+\alpha^2}}\text{ so }\sin2\theta=2\frac{u\alpha}{u^2+\alpha^2}. \]
    • Therefore, the final answer is \[ -\frac{1}{4(2x^2+x+2)}+2^{-3/2}\alpha^{-2}\left(\tan^{-1}\frac{\sqrt2x+2^{-3/2}}{\alpha}+\frac{(\sqrt2x+2^{-3/2})\alpha}{(\sqrt2+2^{-3/2})^2+\alpha^2}\right)+C. \]
  • Exercises. 7.4: 35, 37.

Rationalizing substitutions

  • When an integrand contains an expression of the form \(\sqrt[n]{g(x)}\), then the substitution \(u=\sqrt[n]{g(x)}\) may be effective.
  • \(\int\frac{\sqrt{x-3}}{x}\,\mathrm dx\).
    • Let \(u=\sqrt{x-3}\). Then we have \(u^2+3=x\), and \(2\sqrt{x-3}\,\mathrm du=\mathrm dx\).
    • Therefore, applying this substitution to the integral, we get \[ 2\int\frac{u^2}{u^2+3}\,\mathrm du=2\left(u+\frac{1}{\sqrt3}\tan^{-1}\frac{u}{\sqrt3}\right)+C=2\left(\sqrt{x-3}+\frac{1}{\sqrt3}\tan^{-1}\frac{\sqrt{x-3}}{\sqrt3}\right)+C. \]
  • Exercises. 7.4: 39, 41, 47, 49.