Question

How do you integrate $\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}$ using partial fractions?

Answer 1

Here in this question, we have to find the integrated value of a given function by using partial fractions method. I would try Integration by Partial Fractions when an integrand is a rational function with its denominator can be factored out into smaller factors and by further simplifying using the standard integration formula. To get the required solution.

Complete step by step solution:
If the integrand (the expression after the integral sign) is in the form of an algebraic fraction and the integral cannot be evaluated by simple methods, the fraction needs to be expressed in partial fractions before integration takes place.
The steps needed to decompose an algebraic fraction into its partial fractions results from a consideration of the reverse process − addition (or subtraction).
Consider the given algebraic fractions:
$\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}$---------(1)
Separate the fraction that we wish to decompose in to multiple fractions. The factor of $x$ in the denominator has a power higher than 1, then the coefficients in the numerator should reflect this higher power.
To use the Method of Partial Fractions, we let, for, $\;A,B,C \in R,:$
Equation (1) can be written as
$\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{A}{{x - 1}} + \dfrac{{Bx + C}}{{{x^2} + 4}}$--------(2)
Take $\left( {x - 1} \right)\left( {{x^2} + 4} \right)$ as LCM in RHS, then
$\Rightarrow \,\,\,x = A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)$----------(3)
$A$ can easily be obtained by using Heavyside's Cover-up Method, The Heaviside cover-up method, is one possible approach in determining the coefficients when performing the partial-fraction expansion of a rational function.
To find $A$, Put $x = 1$in equation (3)
$\Rightarrow \,\,\,1 = A\left( {{1^2} + 4} \right) + \left( {B\left( 1 \right) + C} \right)\left( {1 - 1} \right)$
$\Rightarrow \,\,\,1 = A\left( {1 + 4} \right) + \left( {B + C} \right)\left( 0 \right)$
$\Rightarrow \,\,\,1 = A\left( 5 \right) + 0$
Divide both side by 5, then
$\therefore \,\,\,A = \dfrac{1}{5}$
To find $C$, with the value of $A$ and put $x = 0$ in equation (3)
$\Rightarrow \,\,\,x = A\left( {{x^2} + 4} \right) + \left( {Bx + C} \right)\left( {x - 1} \right)$
$\Rightarrow \,\,\,0 = \left( {\dfrac{1}{5}} \right)\left( {{0^2} + 4} \right) + \left( {B\left( 0 \right) + C} \right)\left( {0 - 1} \right)$
$\Rightarrow \,\,\,0 = \left( {\dfrac{1}{5}} \right)4 + \left( {0 + C} \right)\left( { - 1} \right)$
$\Rightarrow \,\,\,0 = \left( {\dfrac{1}{5}} \right)4 - C$
Add both side by $C$, then
$\therefore \,\,\,C = \dfrac{4}{5}$
Substitute the value of $A$ and $C$ in equation (3), then
$\Rightarrow \,\,\,x = \left( {\dfrac{1}{5}} \right)\left( {{x^2} + 4} \right) + \left( {Bx + \dfrac{4}{5}} \right)\left( {x - 1} \right)$
$\Rightarrow \,\,\,x = \left( {\dfrac{1}{5}} \right){x^2} + \left( {\dfrac{1}{5}} \right)4 + B{x^2} - Bx + \dfrac{4}{5}x - \dfrac{4}{5}$
To find $B$ by comparing the ${x^2}$ co-efficient on both side,
$\Rightarrow \,\,\,0 = \dfrac{1}{5} + B$
Subtract $\dfrac{1}{5}$ on both side, then
$\therefore \,\,\,B = - \dfrac{1}{5}$
Substitute the value of $A,B,$ and $C$ in equation (2):
$\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{{\left( {\dfrac{1}{5}} \right)}}{{x - 1}} + \dfrac{{\left( { - \dfrac{1}{5}} \right)x + \left( {\dfrac{4}{5}} \right)}}{{{x^2} + 4}}$
It can be written as
$\Rightarrow \,\,\,\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}} = \dfrac{1}{{5\left( {x - 1} \right)}} - \dfrac{x}{{5\left( {{x^2} + 4} \right)}} + \dfrac{4}{{5\left( {{x^2} + 4} \right)}}$
Integrate each fraction with respect to $x$
$\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}} \,dx = \int {\dfrac{1}{{5\left( {x - 1} \right)}}\,dx} - \int {\dfrac{x}{{5\left( {{x^2} + 4} \right)}}\,dx} + \int {\dfrac{4}{{5\left( {{x^2} + 4} \right)}}} \,dx$----(4)
By integrating $\Rightarrow \,\,\,\int {\dfrac{1}{{5\left( {x - 1} \right)}}\,dx}$ we get
$\Rightarrow \,\,\,\dfrac{1}{5}\int {\dfrac{1}{{\left( {x - 1} \right)}}\,dx}$
$\Rightarrow \,\,\,\ln \left| {x - 1} \right| + k$-------(a)
Next, integrate $\Rightarrow \,\,\, - \int {\dfrac{x}{{5\left( {{x^2} + 4} \right)}}\,dx}$, we get
Multiply both numerator and denominator by 2
$\Rightarrow \,\,\, - \int {\dfrac{{2x}}{{10\left( {{x^2} + 4} \right)}}\,dx}$
Put ${x^2} + 4 = t$
Differentiate w.r.t $x$ $\Rightarrow \,\,\,2x\,dx = dt$, then
$\Rightarrow \,\,\, - \dfrac{1}{{10}}\int {\dfrac{{dt}}{{\left( t \right)}}}$
$\Rightarrow \,\,\, - \dfrac{1}{{10}}\ln |t| + k$
$\Rightarrow \,\,\, - \dfrac{1}{{10}}\ln |{x^2} + 4| + k$------(b)
Next, integrate $\Rightarrow \,\,\,\int {\dfrac{4}{{5\left( {{x^2} + 4} \right)}}} \,dx$, we get
$\Rightarrow \,\,\,\dfrac{4}{5}\int {\dfrac{1}{{\left( {{x^2} + {2^2}} \right)}}} \,dx$
Then by the integrating formula of ${\tan ^{ - 1}}x$, then
$\Rightarrow \,\,\,\dfrac{4}{5}.\dfrac{1}{2}{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right) + k$
$\Rightarrow \,\,\,\dfrac{2}{5}{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right) + k$------(C)

Hence, substitute the (a), (b) and (c) in equation (4) to get the required solution
$\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}} \,dx = \ln \left| {x - 1} \right| + k - \dfrac{1}{{10}}\ln |{x^2} + 4| + k + \,\dfrac{2}{5}{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right) + k$
$\Rightarrow \,\,\,\int {\dfrac{x}{{\left( {x - 1} \right)\left( {{x^2} + 4} \right)}}} \,dx = \ln \left| {x - 1} \right| - \dfrac{1}{{10}}\ln |{x^2} + 4| + \,\dfrac{2}{5}{\tan ^{ - 1}}\left( {\dfrac{x}{2}} \right) + K$
Where $K$ is an integrating constant.

Note: When the function is in the form of a fraction where the denominator of the function is in the form of polynomial, then we use the partial fractions and then we integrate the function. The integration is a reciprocal of the differentiation and by using the standard formulas of integration we determine the values.