Integration of Rational Fractions

To evaluate the integration of an algebraic rational fraction, its rational fraction must be resolved into partial fractions. Often, the denominator of the given rational fraction can be resolved into real linear and quadratic factors which may be single or repeated. According to the nature of the factors of the denominator, we have to adopt different procedures for the breaking up of the rational fraction into partial fractions. Some examples of the integration of rational fractions are given below.


$\int\frac{13}{(3x+4)(4x+1)}dx$

\[\text{Let }\frac{13}{(3x+4)(4x+1)}=\frac{A}{(3x+4)}+\frac{B}{(4x+1)}\text{ __(1)}\] \[13=A(4x+1)+B(3x+4)\] \[13=4Ax+A+3Bx+4B\] \[13=(4A+3B)x+(A+4B)\] \[\therefore 4A+3B=0\text{ and }A+4B=13\] Solving for $A$ and $B$, we get $A=-3$ and $B=4$. Putting the values of $A$ and $B$ in $\text{(1)}$, \[\frac{13}{(3x+4)(4x+1)}=\frac{-3}{(3x+4)}+\frac{4}{(4x+1)}\] \[\therefore \int\frac{13}{(3x+4)(4x+1)}dx=-\int\frac{3dx}{3x+4}+\int\frac{4dx}{4x+1}\] \[=-\log(3x+4)+\log(4x+1)+C\] \[=\log(4x+1)-\log(3x+4)+C\]


$\int\frac{x+1}{x^2-2x-8}dx$

\[\int\frac{x+1}{x^2-2x-8}dx=\int\frac{x+1}{(x+2)(x-4)}dx\] \[\text{Let }\frac{x+1}{(x+2)(x-4)}=\frac{A}{x+2}+\frac{B}{x-4}\text{ __(1)}\] \[x+1=A(x-4)+B(x+2)\] \[x+1=Ax-4A+Bx+2B\] \[x+1=(A+B)x+(2B-4A)\] \[\therefore A+B=1\text{ and }2B-4A=1\] Solving for $A$ and $B$, we get $A=\frac{1}{6}$ and $B=\frac{5}{6}$. Putting the values of $A$ and $B$ in $\text{(1)}$, \[\frac{x+1}{(x+2)(x-4)}=\frac{1}{6(x+2)}+\frac{5}{6(x-4)}\] \[\therefore\int\frac{(x+1)}{x^2-2x-8}dx=\frac{1}{6}\int\frac{dx}{x+2}+\frac{5}{6}\int\frac{dx}{x-4}\] \[=\frac{1}{6}\left[\log(x+2)+5\log(x-4)\right]+C\]


The integration of a rational fraction in which the numerator and the denominator consist of only even powers of $x$ is illustrated below.

$\int\frac{x^2}{x^4-2x^2-15}dx$

Put $x^2=y$, then, \[\frac{x^2}{x^4-2x^2-15}=\frac{y}{y^2-2y-15}=\frac{y}{(y+3)(y-5)}\] \[\text{Let }\frac{y}{(y+3)(y-5)}=\frac{A}{y+3}+\frac{B}{y-5}\text{ __(1)}\] \[y=A(y-5)+B(y+3)\] \[y=Ay-5A+By+3B\] \[y=(A+B)y+(3B-5A)\] \[\therefore A+B=1\text{ and }3B-5A=0\] Solving for $A$ and $B$, we get $A=\frac{3}{8}$ and $B=\frac{5}{8}$. Putting the values of $A$ and $B$ in equation $\text{(1)}$, \[\frac{y}{(y+3)(y-5)}=\frac{3}{8(y+3)}+\frac{5}{8(y-5)}\] \[\therefore\frac{x^2}{x^4-2x^2-15}=\frac{3}{8(x^2+3)}+\frac{5}{8(x^2-5)}\] \[\therefore\int\frac{x^2}{x^4-2x^2-15}dx=\frac{3}{8}\int\frac{dx}{x^2+3}+\frac{5}{8}\int\frac{dx}{x^2-5}\] \[=\frac{3}{8\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+\frac{5}{8}.\frac{1}{2\sqrt{5}}\log\frac{x-\sqrt{5}}{x+\sqrt{5}}+C\] \[=\frac{\sqrt{3}}{8}\tan^{-1}\frac{x}{\sqrt{3}}+\frac{\sqrt{5}}{16}\log\frac{x-\sqrt{5}}{x+\sqrt{5}}+C\]


The integration of a rational fraction in which the numerator consists of only odd powers of $x$ and the denominator only even powers of $x$ is illustrated below.

$\int\frac{x^3}{2x^4-3x^2-5}dx$

\[\int\frac{x^3}{2x^4-3x^2-5}dx=\int\frac{x^2.x}{2x^4-3x^2-5}dx\] Put $x^2=y$, then, $2xdx=dy\Rightarrow xdx=\frac{1}{2}dy$. \[\int\frac{x^2.xdx}{2x^4-3x^2-5}=\int\frac{y.\frac{1}{2}dy}{2y^2-3y-5}\] \[=\frac{1}{2}\int\frac{y}{(2y-5)(y+1)}dy\] \[\text{Let }\frac{y}{(2y-5)(y+1)}=\frac{A}{2y-5}+\frac{B}{y+1}\text{ __(1)}\] \[y=A(y+1)+B(2y-5)\] \[y=Ay+A+2By-5B\] \[y=(A+2B)y+(A-5B)\] \[\therefore A+2B=1\text{ and }A-5B=0\] Solving for $A$ and $B$, we get $A=\frac{5}{7}$ and $B=\frac{1}{7}$. Putting the values of $A$ and $B$ in equation $\text{(1)}$, \[\frac{y}{(2y-5)(y+1)}=\frac{5}{7(2y-5)}+\frac{1}{7(y+1)}\] \[\therefore\frac{1}{2}\int\frac{y}{(2y-5)(y+1)}dy\]\[=\frac{1}{2}\left[\int\frac{5dy}{7(2y-5)}+\int\frac{dy}{7(y+1)}\right]\] \[=\frac{1}{14}\left[\frac{5}{2}\log(2y-5)+\log(y+1)\right]+C\] \[=\frac{1}{14}\left[\frac{5}{2}\log(2x^2-5)+\log(x^2+1)\right]+C\]


For any real numbers $a$ and $b$ $(a≠b)$ and positive integers $m$ and $n$, \[\int\frac{dx}{(x-a)^m(x-b)^n}\] is the other case which can be integrated as illustrated below.

$\int\frac{dx}{(x-1)^2(x-2)^3}$

Put $x-1=z(x-2)$ so that $x=\frac{2z-1}{z-1}$. \[\therefore dx=\frac{2(z-1)-(2z-1)}{(z-1)^2}dz=-\frac{1}{(z-1)^2}dz\] Now, \[\frac{1}{(x-1)^2(x-2)^3}=\frac{1}{z^2(x-2)^5}=\frac{1}{z^2\left(\frac{2z-1}{z-1}-2\right)^5}\] \[=\frac{1}{z^2\left(\frac{2z-1-2z+2}{z-1}\right)^5}=\frac{(z-1)^5}{z^2}\] \[\therefore \int\frac{dx}{(x-1)^2(x-2)^3}=\int\frac{(z-1)^5}{z^2}×\frac{-1}{(z-1)^2}dz\] \[=-\int\frac{(z-1)^3}{z^2}dz=-\int\frac{z^3-3z^2+3z-1}{z^2}dz\] \[=-\int\left(z-3+\frac{3}{z}-\frac{1}{z^2}\right)dz\] \[=-\frac{z^2}{2}+3z-3\log z-\frac{1}{z}+C\] \[=-\frac{1}{2}\left(\frac{x-1}{x-2}\right)^2+3\left(\frac{x-1}{x-2}\right)-3\log\left(\frac{x-1}{x-2}\right)\]\[-\left(\frac{x-2}{x-1}\right)+C\]

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