Question

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

Answer 1

Hint : 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:
$\dfrac{1}{{{{\left( {s + 1} \right)}^2}}}$ ---------(1)
Separate the fraction that we wish to decompose in to multiple fractions. The factor of $s$ 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
$\Rightarrow \dfrac{1}{{{{\left( {s + 1} \right)}^2}}} = \dfrac{A}{{\left( {s + 1} \right)}} + \dfrac{B}{{{{\left( {s + 1} \right)}^2}}}$ --------(2)
Take ${\left( {s + 1} \right)^2}$ as LCM in RHS, then
$\Rightarrow \dfrac{1}{{{{\left( {s + 1} \right)}^2}}} = \dfrac{{A\left( {s + 1} \right) + B}}{{{{\left( {s + 1} \right)}^2}}}$
On simplification, we get
$\Rightarrow 1 = A\left( {s + 1} \right) + B$ ----------(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 $B$ , Put $s = - 1$ in equation (3)
$\Rightarrow 1 = A\left( { - 1 + 1} \right) + B$
$\Rightarrow 1 = A\left( 0 \right) + B$
$\Rightarrow 1 = 0 + B$
$\therefore B = 1$
To find $A$ by comparing the ‘ $s$ ’ co-efficient on both side,
$\Rightarrow 0 = A$
$\therefore A = 0$
Substitute the value of $A$ and $B$ in equation (2):
$\Rightarrow \dfrac{1}{{{{\left( {s + 1} \right)}^2}}} = \dfrac{0}{{\left( {s + 1} \right)}} + \dfrac{1}{{{{\left( {s + 1} \right)}^2}}}$
On simplification, we get
$\Rightarrow \dfrac{1}{{{{\left( {s + 1} \right)}^2}}} = 0 + \dfrac{1}{{{{\left( {s + 1} \right)}^2}}}$
$\Rightarrow \dfrac{1}{{{{\left( {s + 1} \right)}^2}}} = \dfrac{1}{{{{\left( {s + 1} \right)}^2}}}$
Integrate both sides with respect to $x$ .
$\Rightarrow \int {\dfrac{1}{{{{\left( {s + 1} \right)}^2}}}ds} = \int {\dfrac{1}{{{{\left( {s + 1} \right)}^2}}}ds}$ ----(4)
on integrating
$\Rightarrow \int {\dfrac{1}{{{{\left( {s + 1} \right)}^2}}}ds}$
Let $u = s + 1$ , then $du = ds$ and the integral becomes:
$\Rightarrow \int {\dfrac{1}{{{u^2}}}du}$
or
$\Rightarrow \int {{u^{ - 2}}du}$ -------(5)
As we know the standard integration formula $\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$ , then
On integrating equation (5), we get
$\Rightarrow \dfrac{{{u^{ - 2 + 1}}}}{{ - 2 + 1}} + c$
$\Rightarrow \dfrac{{{u^{ - 1}}}}{{ - 1}} + c$
Substitute $u = s + 1$ , then
$\Rightarrow - {\left( {s + 1} \right)^{ - 1}} + c$
Or
$\Rightarrow - \dfrac{1}{{\left( {s + 1} \right)}} + c$
Hence, it’s a required solution.
So, the correct answer is “ $- \dfrac{1}{{\left( {s + 1} \right)}} + c$ ”.

Note : When the function is in the form of a fraction where the denominator of the function is in the form of a 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.