Integration of Rational Functions

We are very lucky to have \[\frac1{x^2-1}=\frac12\left(\frac1{x-1}-\frac1{x+1}\right).\] Now the integral becomes \begin{eqnarray*} \int \frac{1}{x^2-1}\,dx&=&\frac12\int\frac1{x-1}\,dx-\frac12\int\frac1{x+1}\,dx=\frac12\ln|x-1|-\frac12\ln|x+1|+C. \end{eqnarray*}

Let us find real numbers \(a\), \(b\), and \(c\) that satisfy: \[\frac{11x+15}{x(x-3)(x+5)}=\frac{a}{x}+\frac b{x-3}+\frac c{x+5}.\] Multiplying both sides by \(x(x-3)(x+5)\) yields: \[11x+15=a(x-3)(x+5)+bx(x+5)+cx(x-3).\] We need to find \(a\), \(b\), and \(c\) such that the previous expression holds for every \(x\). There are two ways to find the coefficients.

• First way: Simplify the polynomial on the right. After simplifying, the polynomial becomes: \[11x+15=(a+b+c)x^2+ (2a+5b-3c)x-15a.\] Since these two polynomials have to be the same for all \(x\), we obtain the system of equations: \begin{eqnarray*} a+b+c&=&0\\ 2a+5b-3c&=&11\\ -15a&=&15. \end{eqnarray*} We first solve the last equation for \(a\) and obtain \(a=-1\). Multiplying the first equation by \(3\) and adding it to the second yields \(5a+8b=11\), which together with \(a=-1\) gives us \(8b=11-5a=16\), or, equivalently, \(b=2\). Now from the first equation we obtain \(c=-1\).

• Second way: Substituting values for \(x\). Since the equation \[11x+15=a(x-3)(x+5)+bx(x+5)+cx(x-3)\] must hold for all \(x\), it must hold for \(x=0\), \(x=3\), and \(x=-5\). Substituting these three values into the equation will give us a lot of information about \(a\), \(b\), and \(c\). Setting \(x=0\) gives us \(15=a\cdot (-3)\cdot 5\), which means that \(a=-1\). Substituting \(x=3\) gives us \(48=b\cdot 24\), or, \(b=2\). Substituting \(x=-5\) gives us \(-40=40c\), hence \(c=-1\)

Hence we have \[\frac{11x+15}{x(x-3)(x+5)}=\frac{-1}{x}+\frac2{x-3}+\frac{-1}{x+5}.\] We can now evaluate the integral easily: \[\int \frac{11x+15}{x(x-3)(x+5)}\,dx=-\int\frac1x\,dx+2\int\frac1{x-3}\,dx-\int\frac1{x+5}\,dx=-\ln|x|+2\ln|x-3|-\ln|x+5|+C.\]

The original fraction can be represented in the form: \[ \frac{3x^2 +18x +29}{(x +2)^2(x +3)}=\frac{\alpha}{x +2}+\frac{\beta}{(x+2)^2}+\frac{\gamma}{(x+3)}.\] Multiplying both sides of the previous equation by \( (x +2)^2(x +3) \) implies \[ 3x^2 +18x +29=\alpha (x +2)(x +3)+ \beta (x +3)+\gamma (x +2)^2.\quad\quad\quad\quad\quad (1)\]

Substituting \( x=-2 \) in yields (1) \( 12 -36 +29= \beta (-2 +3) \), which gives us \(\beta=5\).

Substituting \( x=-3\) yields \(27 -54 +29= \gamma (-3 +2)^2 \), which gives us \(\gamma=2\).

We can find \(\alpha\) by replacing \(x\) in (1) with some new value different from \( -2\) and \( -3 \). For example, setting \( x=0 \) yields \[ +29= \alpha\cdot (0 +2)(0 +3) + 5\cdot (0 +3) + 2\cdot (0 +2)^2,\] giving us \(\alpha={1}\).

There is a polynomial \(P\) and real numbers \(\alpha\), \(\beta\), and \(\gamma\) such that the original fraction can be represented in the form: \[ \frac{3x^4-9x^3+20x^2-43x+14}{\left(x^2+4\right)(x -3)}=P(x)+\frac{\alpha x+\beta}{x^2+ 4}+\frac{\gamma}{x-3}.\] We first find divide the polynomial \(3x^4-9x^3+20x^2-43x+14\) by \(\left(x^2+4\right)(x -3)=x^3-3x^2+4x-12\) to obtain the quotient \(3x\) and the remainder \(8x^2-7x+14\). Thus: \[ \frac{3x^4-9x^3+20x^2-43x+14}{\left(x^2+4\right)(x -3)}=3x+\frac{8x^2-7x+14}{\left(x^2+4\right)(x -3)}.\] We now want to find \(\alpha\), \(\beta\), and \(\gamma\) such that: \[ \frac{8x^2 -7x +14}{\left(x^2+4\right)(x -3)}=\frac{\alpha x+\beta}{x^2+ 4}+\frac{\gamma}{x-3}.\] If we multiply both sides of the previous equation by \( \left(x^2+4\right)(x -3) \) the equation becomes equivalent to \[ 8x^2 -7x +14=(\alpha x+\beta)(x-3)+\gamma(x^2+4).\quad\quad\quad\quad\quad (1)\]

Substituting \( x=3 \) in (1) yields \( 65= \gamma (3^2 +4) \), which gives us \(\gamma=5\).

By expanding the equation (1) and substituting \(\gamma=5\) we get: \[8x^2 -7x +14=(\alpha+5)x^2+(\beta -3\alpha)x -3\beta+5\cdot 4.\] This gives us \(8=\alpha+5\) which implies \(\alpha=3 \). From the equation \(-7=\beta-3\cdot \alpha \) and \(\alpha=3\) we obtain \(\beta=2\). Thus \[\frac{3x^4-9x^3+20x^2-43x+14}{\left(x^2+4\right)(x -3)}=3x+\frac{3 x+2}{x^2+ 4}+\frac{5}{x-3}.\]

Notice that \(\displaystyle \frac1{(x^2-a^2)^2}=\frac1{(x-a)^2(x+a)^2}\). Let us find the real numbers \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\) such that \[\frac1{(x-a)^2(x+a)^2}=\frac{\alpha}{x-a}+\frac{\beta}{(x-a)^2}+\frac{\gamma}{x+a}+\frac{\delta}{(x+a)^2}.\] Multiplying both sides of the previous equation by \((x-a)^2(x+a)^2\) we get: \[1=\alpha(x-a)(x+a)^2+\beta(x+a)^2+\gamma(x-a)^2(x+a)+\delta (x-a)^2. \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad (1)\] Plugging \(x=a\) in the previous expression yields \(\beta=\frac1{4a^2}\). Plugging \(x=-a\) in (1) yields \(\delta=\frac1{4a^2}\). Since (1) holds for all \(x\), the derivative of the left side of (1) should be equal to the derivative of the right side of (1). Thus: \[0=\alpha\left((x+a)^2+2(x-a)(x+a)\right)+ 2\beta(x+a)+\gamma\left(2(x-a)(x+a)+(x-a)^2\right)+2\delta (x-a). \quad\quad\quad (1)\] Setting \(x=a\) in (2) yields: \( 4a^2\alpha+\frac1a=0\) hence \(\alpha=-\frac1{4a^3}\). Similarly, setting \(x=-a\) in (2) yields \(\gamma=\frac1{4a^3}\). Thus \begin{eqnarray*} \int\frac1{(x^2-a^2)^2}\,dx&=&-\frac1{4a^3}\int\frac1{x-a}\,dx+\frac1{4a^2}\frac1{(x-a)^2}\,dx+\frac1{4a^3}\int \frac1{x+a}\,dx+\frac1{4a^2}\int\frac1{(x+a)^2}\,dx\\ &=&\frac1{4a^3}\ln\left|\frac{x+a}{x-a}\right|-\frac1{4a^2}\left(\frac1{x-a}+\frac1{x+a}\right)+C. \end{eqnarray*}

You may omit the proof of this theorem and you will still be able to learn the technique of partial fractions. This proof is presented here for completeness, and is a fairly easy consequence of Fundamental Theorem of Algebra and of the Euclidean algorithm for polynomials.

According to Fundamental Theorem of algebra the polynomial \(B\) can be factored in linear and quadratic terms. Hence there are real number \(K\), real numbers \(p\) and \(q\), sequences of real numbers \((\alpha_i)_{i=1}^p\), \((\beta_i)_{i=1}^q\), \((\gamma_i)_{i=1}^q\), and sequences of positive integers \((s_i)_{i=1}^p\) and \((t_i)_{i=1}^q\) such that all \(\alpha_i\) are distinct, all pairs \((\beta_i,\gamma_i)\) are distinct, and \[B(x)=K\cdot \prod_{i=1}^p (x+\alpha_i)^{s_i}\cdot \prod_{i=1}^q \left((x+\gamma_i)^2+\beta_i^2\right)^{t_i}.\] The factors \((x+\alpha_i)^{s_i}\), \left((x+\gamma_j)^2+\beta^j\right)^{t_j} are mutually relatively prime polynomials. If \(P(x)\) and \(Q(x)\) are relatively prime then according to euclidean algorithm applied to polynomials there are polynomials \(U\) and \(V\) such that \(1=U(x)P(x)+V(x)Q(x)\). This is equivalent to \[\frac1{P(x)Q(x)}=\frac{V(x)}{P(x)}+\frac{U(x)}{Q(x)}.\] Using induction we get a sequence of polynomials \(U_1(x)\), \(\dots\), \(U_{p}(x)\), \(V_1(x)\), \(\dots\), \(V_q(x)\) such that \[\frac1{B(x)}=\sum_{i=1}^{p}\frac{U_i(x)}{(x+\alpha_i)^{s_i}}+\sum_{i=1}^q\frac{V_i(x)}{\left((x+\gamma_i)^2+\beta_i\right)^{t_i}}.\] Therefore we have \[\frac{A(x)}{B(x)}=\sum_{i=1}^{p}\frac{A(x)U_i(x)}{(x+\alpha_i)^{s_i}}+\sum_{i=1}^q\frac{A(x)V_i(x)}{\left((x+\gamma_i)^2+\beta_i\right)^{t_i}}.\] It remains to prove the following two propositions:

• Proposition 1. Each of the fractions \(\displaystyle \frac{A(x)U_i(x)}{(x+\alpha_i)^{s_i}}\) can be represented as a sum of a polynomial and summands of the form \(\displaystyle \frac{d_i}{(x+\alpha_i)^k}\)

• Proposition 2. Each of the fractions \(\displaystyle \frac{A(x)V_i(x)}{\left((x+\gamma_i)^2+\beta_i\right)^{t_i}}\) can be expressed as a sum of a polynomials and summands of the form \(\displaystyle \frac{e_ix+f_i}{\left((x+\gamma_i)^2+\beta_i^2\right)^k}\).

The first proposition is proved by introducing the substitution \(y=x+\alpha_i\). We can rewrite \(A(x)U_i(x)=S_i(y)\) for a polynomial \[S_i(y)=q_{i,k_i}y^{k_i}+q_{i,k_i-1}y^{k_i-1}+\cdots + q_{i,0},\] giving us: \[\frac{A(x)U_i(x)}{(x+\alpha_i)^{s_i}}=\frac{S_i(y)}{y^{s_i}}= q_{i,k_i}y^{k_i-s_i}+ q_{i,k_i-1}y^{k_i-s_i-1}+\cdots +q_{i,0}y^{-s_i}\] which is exactly what we wanted to prove.

The second proposition is proved in a similar way. We introduce the substitution \(y=x+\gamma_i\). There exists a polynomial \(T_i\) such that \(A(x)V_i(x)=T_i(y)\). The polynomial \(T_i(y)\) can be divided by \(\left(y^2+\beta_i^2\right)^{t_i}\) with remainder giving us \(T_i(y)=\left(y^2+\beta_i^2\right)^{t_i} Q_{i,1}(y)+ T_{i,1}(y)\), for some polynomials \(Q_{i,1}(y)\) and \(T_{i,1}(y)\) such that \(\deg T_{i,1} < 2t_i\). There are polynomials \(Q_{i,2}(y)\) and \(T_{i,2}(y)\) such that \(\deg T_{i,2} < 2(t_i-1)\) and \(T_{i,1}(y)=\left(y^2+\beta_i^2\right)^{t_i-1}Q_{i,2}(y)+ T_{i,2}(y)\). Since the degree of the polynomial \(T_{i,1}\) was at most \(2t_i-1\) we see that the degree of \(Q_{i,2}\) is at most 1. Continuing by induction we get a sequence of polynomials \(Q_{i,1}\), \(Q_{i,2}\), \(\dots\), \(Q_{i,t_i}\) such that \(\deg Q_{i,j}\leq 1\) for all \(j\geq 2\) and \[T_i(y)=\left(y^2+\beta_i^2\right)^{t_i} Q_{i,1}(y)+\left(y^2+\beta_i^2\right)^{t_i-1} Q_{i,2}(y)+ Q_{i,t_i+1}(y),\] and degree of each of \(Q_{i,j}\) is at most 1 for \(j\geq 2\). Dividing both sides by \(\displaystyle \left(y^2+\beta_i^2\right)^{t_i} \) yields \[\frac{T_i(y)}{\left(y^2+\beta_i^2\right)^{t_i} }=Q_{i,1}(y)+\frac{Q_{i,2}(y)}{y^2+\beta_i^2}+\frac{Q_{i,3}(y)}{\left(y^2+\beta_i^2\right)^{2}}+\cdots + \frac{Q_{i,t_i+1}(y)}{\left(y^2+\beta_i^2\right)^{t_i} }.\] This proves Proposition 2 and completes the proof of the theorem.

Consider the integral \(\displaystyle \int\frac1{ x^2+b^2}\,dx\) that we already know how to evaluate. Let us try to evaluate the very same integral using the integration by parts with \(\displaystyle u=\frac1{ x^2+b^2}\) and \(v=x\). We will get stuck, but we will accidentally evaluate the integral \(\displaystyle \int\frac1{\left( x^2+b^2\right)^2}\,dx\).

We first evaluate \(\displaystyle du=-\frac{2 x}{\left( x^2+b^2\right)^2}\) and then \begin{eqnarray*}\int\frac1{ x^2+b^2}\,dx &=& \frac{x}{ x^2+b^2}+\int \frac{2 x^2}{\left( x^2+b^2\right)^2}\,dx\\ &=& \frac{x}{ x^2+b^2}+2\int \frac{ x^2+b^2-b^2}{\left( x^2+b^2\right)^2}\,dx \\&=& \frac{x}{ x^2+b^2}+2\int\frac1{ x^2+b^2}\,dx-2b^2\int\frac1{\left( x^2+b^2\right)^2}\,dx. \end{eqnarray*} Thus: \begin{eqnarray*} \int\frac1{\left( x^2+b^2\right)^2}\,dx&=&\frac1{2b^2}\left(\frac{x}{ x^2+b^2}+\int\frac1{ x^2+b^2}\,dx \right)\\ &=& \frac1{2b^2}\left(\frac{x}{ x^2+b^2}+\frac1{b}\tan^{-1}\left(\frac{x}b\right) \right)+C. \end{eqnarray*}

We will use the same strategy as in the proof of Theorem 5. We will use failed integration by parts applied to the integral \(\displaystyle \int\frac1{\left( x^2+b^2\right)^2}\,dx\) that we already know how to evaluate. Let us denote \(\displaystyle u=\frac1{\left( x^2+b^2\right)^2}\) and \(v=x\).

We first evaluate \(\displaystyle du=-\frac{4 x}{\left( x^2+b^2\right)^3}\) and then \begin{eqnarray*}\int\frac1{\left( x^2+b^2\right)^2}\,dx &=& \frac{x}{\left( x^2+b^2\right)^2}+\int \frac{4 x^2}{\left( x^2+b^2\right)^3}\,dx\\ &=& \frac{x}{\left( x^2+b^2\right)^2}+4\int \frac{ x^2+b^2-b^2}{\left( x^2+b^2\right)^3}\,dx \\&=& \frac{x}{\left( x^2+b^2\right)^2}+4\int\frac1{\left( x^2+b^2\right)^2}\,dx-4b^2\int\frac1{\left( x^2+b^2\right)^3}\,dx. \end{eqnarray*} Thus: \begin{eqnarray*} \int\frac1{\left( x^2+b^2\right)^3}\,dx&=&\frac1{4b^2}\left(\frac{x}{\left( x^2+b^2\right)^2}+3\int\frac1{\left( x^2+b^2\right)^2}\,dx \right)\\ &=& \frac1{4b^2}\left(\frac{x}{\left( x^2+b^2\right)^2}+\frac3{2b^2}\left(\frac{x}{ x^2+b^2}+\frac1{b}\tan^{-1}\left(\frac{x}b\right) \right) \right)+C\\ &=&\frac{3 x^3+5b^2x}{8b^4\left( x^2+b^2\right)^2}+\frac{3\tan^{-1}\left(\frac{x}b\right)}{8b^5}+C. \end{eqnarray*}

We will use the same strategy as in the proof of Theorems 5 and 6. Let us denote \(\displaystyle u=\frac1{\left( x^2+b^2\right)^{k-1}}\) and \(v=x\).

We first evaluate \(\displaystyle du=-\frac{2(k-1) x}{\left( x^2+b^2\right)^k}\) and then \begin{eqnarray*}\int\frac1{\left( x^2+b^2\right)^{k-1}}\,dx &=& \frac{x}{\left( x^2+b^2\right)^{k-1}}+\int \frac{2(k-1) x^2}{\left( x^2+b^2\right)^k}\,dx\\ &=& \frac{x}{\left( x^2+b^2\right)^{k-1}}+2(k-1)\int \frac{ x^2+b^2-b^2}{\left( x^2+b^2\right)^k}\,dx \\&=& \frac{x}{\left( x^2+b^2\right)^{k-1}}+2(k-1)\int\frac1{\left( x^2+b^2\right)^{k-1}}\,dx-2(k-1)b^2\int\frac1{\left( x^2+b^2\right)^k}\,dx. \end{eqnarray*} Thus: \begin{eqnarray*} \int\frac1{\left( x^2+b^2\right)^k}\,dx&=&\frac1{2(k-1)b^2}\left(\frac{x}{\left( x^2+b^2\right)^{k-1}}+(2k-3)\int\frac1{\left( x^2+b^2\right)^{k-1}}\,dx \right). \end{eqnarray*}

Let us use the substitution \(u= x^2+b^2\). Then we have \(du=2 x\,dx\) which implies that \(xdx=\frac1{2 }\,du\). The integral becomes \begin{eqnarray*} \int\frac{x}{ x^2+b^2 }\,dx&=& \int\frac{\frac1{2 }\,du}{u}=\frac1{2 }\cdot\ln|u|+C= \frac{\ln\left( x^2+b^2\right)}{2 } +C. \end{eqnarray*}

Let us use the substitution \(u= x^2+b^2\). Then we have \(du=2 x\,dx\) which implies that \(xdx=\frac1{2 }\,du\). The integral becomes \begin{eqnarray*} \int\frac{x}{\left( x^2+b^2\right)^k}\,dx&=& \int\frac{\frac1{2 }\,du}{u^k}=\frac1{2 }\cdot \frac1{1-k}u^{1-k}+C= \frac{\left( x^2+b^2\right)^{1-k}}{2 (1-k)}+C. \end{eqnarray*}

The original fraction can be represented in the form: \[ \frac{7x^2 -2x +6}{(x -2)^2(x +3)}=\frac{\alpha}{x -2}+\frac{\beta}{(x-2)^2}+\frac{\gamma}{(x+3)},\] hence \(\delta=0\). Multiplying both sides of the previous equation by \((x -2)^2(x +3)\) we get \[7x^2 -2x +6=\alpha (x -2)(x +3)+ \beta (x +3)+\gamma (x -2)^2.\quad\quad\quad\quad\quad (1)\]

Substituting \(x=2\) in (1) yields \(28 -4 +6= \beta (2 +3)\), which gives us \(\beta=6\).

Substituting \(x=-3\) yields \(63 +6 +6= \gamma (-3 -2)^2\), which gives us \(\gamma=3\).

We can find \(\alpha\) by replacing \(x\) in (1) with some new value (different from \(2\) and \(-3\)). For example, setting \(x=0\) yields \[+6= \alpha\cdot (0 -2)(0 +3) + 6\cdot (0 +3) + 3\cdot (0 -2)^2,\] giving us \(\alpha={4}\). Therefore \[\alpha+\beta+\gamma+\delta={4}+6+3+0=13.\]

The original fraction can be represented in the form: \[ \frac{7x^2 -9x +32}{\left(x^2+25\right)(x -3)}=\frac{\alpha x+\beta}{x^2+ 25}+\frac{\gamma}{(x-3)}.\] If we multiply both sides of the previous equation by \(\left(x^2+25\right)(x -3)\) the equation becomes equivalent to \[7x^2 -9x +32=(\alpha x+\beta)(x-3)+\gamma(x^2+25).\quad\quad\quad\quad\quad (1)\]

Substituting \(x=3\) in (1) yields \(68= \gamma (3^2 +25)\), which gives us \(\gamma=2\).

By expanding the equation (1) and substituting \(\gamma=2\) we get: \[7x^2 -9x +32=(\alpha +2)x^2+(\beta -3\alpha)x -3\beta +2\cdot 25.\] This gives us \(7 =\alpha +2\) implying \(\alpha=5\). From the equation \(-9=\beta-3\cdot \alpha\) (and \(\alpha=5\)) we obtain \(\beta=6\). Thus \[\alpha-\beta+\gamma=5-6+2=1.\]

The original fraction can be represented in the form: \[ \frac{ 4}{x^2 -7x +12 }=\frac{ 4}{(x -4) (x -3) }=\frac{\alpha}{x -4}+\frac{\beta}{x -3}.\] We will first find \(\alpha\) and \(\beta\). If we multiply both sides of the previous equation by \((x -4)(x -3)\) the equation becomes equivalent to \[ 4= \alpha(x -3)+ \beta (x -4).\quad\quad\quad\quad\quad (1)\]

Substituting \(x=4\) in (1) yields \(\alpha=4\) and substituting \(x=3\) in (1) yields \(\beta=-4\). The integral now becomes \[\int_{100}^{200}\frac{ 4}{x^2 -7x +12 }\,dx=\int_{100}^{200}\frac{4}{x -4}\,dx+ \int_{100}^{200}\frac{-4}{x -3}\,dx=4 \ln\frac{196}{96} -4\ln \frac{197}{97}.\]

We first represent the integral in the following form: \[\int_0^1\frac{8x^2 +6x +5}{x^2 +2x+2 }\,dx=\int_0^1 8\,dx+\int_0^1\frac{-10x -11}{(x+1)^2+1}\,dx=8+\int_0^1\frac{-10x -11}{(x+1)^2+1}\,dx.\] In the second integral we use the substitution \(u=x+1\). The integral becomes: \begin{eqnarray*}\int_0^1\frac{-10u -1}{u^2+1}\,du&=&\frac{-10}2\int_1^2\frac{2u\,du}{u^2+1} - \int_1^2\frac1{u^2+1}\,du\\ &=& \left.\frac{-10}2\ln\left|u^2+1\right| \right|_1^2 \left.- \tan^{-1} u\right|_1^2\\ &=&-5\ln\frac52 -1\cdot\left(\tan^{-1}2-\frac{\pi}4\right) \end{eqnarray*} The final answer is \[\int_0^1\frac{8x^2 +6x +5}{x^2 +2x+2 }\,dx=-5\ln\frac52 - \left(\tan^{-1}2-\frac{\pi}4\right)+8.\]