Integration by Partial Fractions

INTEGRATION BY PARTIAL FRACTIONS

A function of the form P(x)/Q(x), where P(x) and Q(x) are polynomials, is called a rational function. Consider the rational function $\frac{x+7}{(\text{2x - 3) (3x + 4)}}=\frac{1}{2x-3}\text{ - }\frac{\text{1}}{\text{3x + 4}}$

The two fractions on the RHS are called partial fractions. To integrate the rational function on the LHS, it is enough to integrate the two fractions on the RHS, which are easily integrable. This is known as method of partial fractions. In case the degree of P(x) (numerator) is not less than that of Q(x) (denominator), we carry out the division of P(x) by Q(x) and reduce the degree of the numerator.

In order to write P(x)/Q(x) in partial fractions, first of all we write

Q(x) = (x - a)^{k ...} (x² + $\alpha$x + $\beta$)^r ... where binomials are different, and then set

$\dfrac{P(x)}{\text{Q(x)}}\text{ = }\dfrac{{{\text{A}}_{\text{1}}}\text{ }}{\text{(x-a)}}\text{ + }\dfrac{{{\text{A}}_{\text{2}}}\text{ }}{{{\text{(x-a)}}^{\text{2}}}}\text{ + }...\text{ + }\dfrac{{{\text{A}}_{\text{k}}}}{{{\text{(x-a)}}^{\text{k}}}}+\text{ }\dfrac{{{\text{M}}_{\text{1}}}\text{x + }{{\text{N}}_{\text{1}}}}{{{\text{x}}^{\text{2}}}\,+\,\alpha x\,+\,\beta }\text{ +}\dfrac{{{\text{M}}_{\text{2}}}\text{x + }{{\text{N}}_{\text{2}}}}{{{\left( {{\text{x}}^{\text{2}}}\,+\,\alpha x\,+\,\beta \right)}^{2}}}+...+\text{ }\dfrac{{{\text{M}}_{\text{r}}}\text{x + }{{\text{N}}_{\text{r}}}}{{{\left( {{\text{x}}^{\text{2}}}\,+\,\alpha x\,+\,\beta \right)}^{r}}}\,+...$.

where A₁, A₂, ..., A_k, M₁, M₂, ......, M_r, N₁, N₂, ...... ,N_r are real constants to be determined. These are determined by reducing both sides of the above identity to integral form and equating the coefficients of equal powers of x, which gives a system of linear equations in the coefficient. (This method is called the method of comparison of coefficients). The constants can also be obtained by substituting suitably chosen numerical values of x in both sides of the identity.

Note: Before proceeding to write a rational function as a sum of partial fractions, we should be taken to ascertain that it is either a proper rational fraction or is rewritten as one.

A rational function $\frac{P\left(x\right)}{Q\left(x\right)}$ is proper if the degree of polynomial Q(x) is greater than the degree of the polynomial P(x). In case the degree of P(x) greater than or equal to the degree of Q(x), we first write $\frac{P\left(x\right)}{Q\left(x\right)}=h\left(x\right)+\frac{p\left(x\right)}{Q\left(x\right)}$, where h(x) is a polynomial and p(x) is a polynomial of degree less than the degree of polynomial Q(x).

Illustration -1: Evaluate $\int \frac{\cos xdx}{\left(1+\sin x\right)\left(2+\sin x\right)}$.

Solution: Put sinx = t $\Rightarrow$ cosx dx = dt

$\int \frac{dt}{\left(1+t\right)\left(2+t\right)}$ =$\int \frac{dt}{\left(1+t\right)}-\int \frac{dt}{\left(2+t\right)}$ = log (1+ t) – log (2 +t ) +c

= log$\frac{\left(1+\sin x\right)}{\left(2+\sin x\right)}+c$.

Illustration -2: Evaluate $\int \frac{\text{xdx}}{{\text{x}}^{\text{3}}+1}$.

Solution: Since x³ + 1 = (x + 1) (x² - x + 1) (the second factor is not a product of linear factors), the partial fractions of the given integer will have the form.

$\frac{\text{x}}{{\text{x}}^{\text{3}}\text{ + 1}}\text{ = }\frac{\text{A}}{\text{x + 1}}\text{ + }\frac{\text{Bx + D}}{{\text{x}}^{\text{2}}\text{ - x + 1}}$.

Hence, x = A(x² - x + 1) + (Bx + D)(x + 1) = (A + B)x²+ (-A + B + D)x + (A + D).

Equating the coefficients of equal powers of x, we get A = -1/3, B = 1/3, D = 1/3.

\begin{align}  \text{Thus I = -}\dfrac{\text{1}}{\text{3}}\text{ }\int{\dfrac{\text{dx}}{\text{x + 1}}\,+\,\dfrac{1}{3}\,\int{\dfrac{x\,+\,1}{{{x}^{2}}\text{ - x + 1}}}}\,dx\,\,=\,-\,\dfrac{1}{3}\,\ln \,\left|x\,+1\right|\,+\,\dfrac{1}{3}\,{{I}_{1}}. \\  To\text{ calculate the integral }\,\,{{\text{I}}_{\text{1}}}\text{ =}\int{\dfrac{\text{x + 1}}{{{\text{x}}^{\text{2}}}\text{ - x + 1}}\text{ dx}}, \\\end{align}

express (x + 1) = l(d.c. of x² - x + 1) + m Þ (x + 1) = l(2x - 1) + m = 2xl - l + m

Comparing the coefficients of like powers of x, we get l =1/2 and m=3/2

$\Rightarrow$ ${\text{I}}_1\text{= }\frac{\text{1}}{\text{2}}\int \frac{2x\text{ - 1}}{{\text{x}}^{\text{2}}\text{ - x + 1}}\text{ dx}+\frac{\text{3}}{\text{2}}\int \frac{\text{dx}}{{\text{x}}^{\text{2}}\text{ - x + 1}}$

$\begin{align} & \text{ } \\ & =\dfrac{1}{2}\text{ ln }\left| {{\text{x}}^{\text{2}}}\text{ - x + 1} \right|\text{ + }\dfrac{\text{3}}{\text{2}}\text{ }\left( \dfrac{\text{2}}{\sqrt{\text{3}}} \right)\text{ ta}{{\text{n}}^{\text{-1}}}\text{ }\dfrac{\text{x - 1/2}}{\sqrt{\text{3}}\text{/2}}\text{ + }{{\text{c}}_{\text{1}}}\,\,\,\,\, \\ & =\,\,\,\dfrac{1}{2}\text{ ln }\left| {{\text{x}}^{\text{2}}}\text{ - x + 1} \right|\text{ + }\sqrt{\text{3}}\text{ ta}{{\text{n}}^{\text{-1}}}\text{ }\left( \dfrac{\text{2x - 1}}{\sqrt{\text{3}}} \right)\text{ + }{{\text{c}}_{\text{1}}} \\ & Hence\text{ I = -}\dfrac{\text{1}}{\text{3}}\text{ ln }\left| \text{x + 1} \right|\,+\dfrac{1}{6}\text{ ln }\left| {{\text{x}}^{\text{2}}}\text{ - x + 1} \right|\,\text{ + }\dfrac{\text{1}}{\sqrt{\text{3}}}\text{ ta}{{\text{n}}^{\text{-1}}}\text{ }\left( \dfrac{\text{2x - 1}}{\sqrt{\text{3}}} \right)\,+\,c \\ \end{align}$