Integration by Partial Fractions: Definition, Forms, Solved Examples

Integration by Partial Fractions: Partial fractions are important in determining the production, distribution, and consumption of commodities and services because they are used to break down complex rational equations into simpler ones. Also, they play important role in estimating the velocity and trajectory of an item, predicting the positions of planets, and understanding electromagnetism.

The approach of decomposing and then integrating a rational fraction integral with complex components in the denominator is known as integration by partial fractions. We compute and break down the equation into simpler terms using partial fractions so that we may simply calculate or integrate the result. Let us study more about the various forms and procedures utilized in partial fraction integration.

Partial Fractions

If \(f(x)\) and \(g(x)\) are polynomials, then \(\frac{f(x)}{g(x)}\) defines a rational algebraic function or a rational function of \(x\).

• If degree of \(f(x) \geq\) degree of \(g(x)\), then \(\frac{f(x)}{g(x)}\) is called an improper rational function.

If \(\frac{f(x)}{g(x)}\) is an improper rational function, we divide \(f(x)\) by \(g(x)\) so that the rational function is expressed in the form

\(\phi(x)+\frac{\psi(x)}{g(x)}\)

Learn Everything About Fractions Here

Where \(\phi(x)\) and \(\psi \left( x \right)\) are polynomials such that the degree of \(\psi(x)\) is less than that of \(g(x)\). Thus \(\frac{f(x)}{g(x)}\) is expressible as the sum of a polynomial and a proper rational fraction.

• If degree of \(f(x)<\) degree of \(g(x)\) then \(\frac{f(x)}{g(x)}\) is called a proper rational fraction.

Any proper rational function \(\frac{f(x)}{g(x)}\) can be expressed as the sum of two rational functions, each having a simple factor of \( g(x)\). Each such fraction is called a partial fraction. The process of obtaining them is called the resolution or decomposition of \(\frac{f(x)}{g(x)}\) into partial fractions.

The resolution of \(\frac{f(x)}{g(x)}\) into partial fractions depends mainly upon the nature of the factors of \(g(x)\) as discussed below:

Denominator with Non-repeated Linear Factors

Case I : When \(g(x)\) contains non-repeated linear factors only

Let \(g(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)\). Then, we assume that

\(\frac{f(x)}{g(x)}=\frac{A_{1}}{x-a_{1}}+\frac{A_{2}}{x-a_{2}}+\cdots+\frac{A_{\pi}}{x-a_{n}}\)

Where, \(A_{1}, A_{2}, \ldots, A_{n}\) are constants.

Steps to determine the constants:

Step 1: Take the LCM on RHS.

Step 2: Equate the numerator on RHS to the numerator on LHS

Step 3: Substitute, \(x=a_{1}, a_{2 \ldots \ldots,} a_{n }\)

Example: Resolve \(\frac{3 x+2}{x^{3}-6 x^{2}+11 x-6}\) into partial fractions.

Solution: Given \(: \frac{3 x+2}{x^{3}-6 x^{2}+11 x-6}=\frac{3 x+2}{(x-1)(x-2)(x-3)}\)

Let \(\frac{3 x+2}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\)……(i)

Then

\(\Rightarrow \frac{3 x+2}{(x-1)(x-2)(x-3)}=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\)

\(\Rightarrow 3 x+2=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\)……(ii)

Now, \(x-1=0 \Rightarrow x=1\)

Substituting \(x=1\) in equation (ii), we have,

\(5=A(1-2)(1-3)\)

\(\Rightarrow 2 A=5\)

\(\Rightarrow A=\frac{5}{2}\)

Now, \(x-2=0 \Rightarrow x=2\)

Substituting \(x=2\) in equation (ii) we have,

\(8=B(2-1)(2-3)\)

\(\Rightarrow B=-8\)

Substituting \(x=3\) in equation (ii), we have,

\(11=2 C\)

\(\Rightarrow C=\frac{11}{2}\)

Substituting the known values of \(A,B\) and \(C\) in equation (i),

\(\frac{3 x+2}{(x-1)(x-2)(x-3)}=\frac{5}{2(x-1)}-\frac{8}{x-2}+\frac{11}{2(x-3)}\)

Remark: In order to determine the values of constants in the numerator of the partial fraction corresponding to the non-repeated linear factor \(px+q\) then denominator of a rational expression, we may proceed as follows:

Replace \(x=-\frac{q}{p}\) obtained by putting  \(p x+q=0\) everywhere in the given rational expression except in the factor \(p x+q\) itself.

Denominator with Some Repeated Linear Factors

Case II : When \(g(x)\) contains some repeated linear factors and the remaining are non-repeated linear factors

Let \(g(x)=(x-a)^{k}\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{r}\right)\). Then, we assume that

\(\frac{f(x)}{g(x)}=\frac{A_{1}}{x-a}+\frac{A_{2}}{(x-a)^{2}}+\frac{A_{3}}{(x-a)^{3}}+\cdots+\frac{A_{k}}{(x-a)^{k}}+\frac{B_{1}}{x-a_{1}}+\frac{B_{2}}{x-a_{2}}+\cdots+\frac{B_{r}}{\left(x-a_{r}\right)}\)

i.e., corresponding to non-repeating factors we assume as in Case I and for each repeating factor \((x-a)^{k}\) we assume  partial fractions  \(\frac{A_{1}}{x-a}+\frac{A_{2}}{(x-a)^{2}}+\frac{A_{3}}{(x-a)^{3}}+\cdots+\frac{A_{k}}{(x-a)^{k}}\) Where \(A_{1}, A_{2}, \ldots, A_{k}\) are constants.

Now, to determine constants we equate numerators on both sides. Some of the constants are determined by comparing coefficients of equal powers of \(x\) on both sides.

Example: Resolve \(\frac{3 x-2}{(x-1)^{2}(x+1)}\) into partial fractions.

Solution: Let \(\frac{3 x-2}{(x-1)^{2}(x+1)(x+2)}=\frac{A_{1}}{x-1}+\frac{A_{2}}{(x-1)^{2}}+\frac{A_{3}}{x+1}+\frac{A_{4}}{x+2}\)

\( \Rightarrow 3x – 2 = {A_1}(x – 1)(x + 1)(x + 2) + {A_2}(x + 1)(x + 2) + {A_3}{(x – 1)^2}(x + 2) + {A_4}{(x – 1)^2}(x + 1)\) ……..(i)

Substituting \(x-1=0\) or \(x=1\) in equation (i), we have,

\(1=A_{2}(2)(3)\)

\(\therefore A_{2}=\frac{1}{6}\)

Substituting \(x+1=0\) or \(x=-1\) in equation (i) we have,

\(-5=A_{3}(-2)^{2}(1)\)

\(\therefore A_{3}=-\frac{5}{4}\)

Substituting \(x+2=0\) or \(x=-2\) in equation (i) we have,

\(-8=A_{4}(-3)^{2}(-2+1)\)

\(\therefore A_{4}=\frac{8}{9}\)

Now, equating coefficient \(x^{3}\) on both sides, we get,

\(0=A_{1}+A_{3}+A_{4}\)

\(\Rightarrow A_{1}=-A_{3}-A_{4}\)

\(=\frac{5}{4}-\frac{8}{9}\)

\(\therefore \mathrm{A}_{1}=\frac{13}{36}\)

\(\therefore \frac{3 x-2}{(x-1)^{2}(x+1)(x+2)}=\frac{13}{36(x-1)}+\frac{1}{6(x-1)^{2}}-\frac{5}{4(x+1)}+\frac{8}{9(x+2)}\)

Denominator with Irreducible Factors

Case III : When \(g(x)\) contains non-repeated irreducible factors of the form \({\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + bx + c}}\)

Corresponding to each quadratic factor of  \(a x^{2}+b x+c\) we assume partial fraction of the type \(\frac{A x+B}{a x^{2}+b x+c}\) where \(A\) and \(B\) are constants to be determined by comparing coefficients of similar powers of \(x\) in the numerator of both sides. In practice it is advisable to assume partial fractions of the type

\(\frac{A(2 a x+b)}{a x^{2}+b x+c}+\frac{B}{a x^{2}+b x+c}\)

Example: Resolve \(\frac{2 x-1}{(x+1)\left(x^{2}+2\right)}\) into partial fractions.

Ans: Let \(\frac{2 x-1}{(x+1)\left(x^{2}+2\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+2}\), Then

\(\frac{2 x-1}{(x+1)\left(x^{2}+2\right)}=\frac{A\left(x^{2}+2\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+2\right)}\)

\(\Rightarrow 2 x-1=A\left(x^{2}+2\right)+(B x+C)(x+1)\)…..(i)

Now, \(x+1=0 \Rightarrow x=-1\)

So, substituting \(x=-1\) in equation (i), we have

\(-3=3 A\)

\(\Rightarrow A=-1\)

Comparing coefficients of like powers of \(x\) on both sides of (i) we get

\(A+B=0, C+2 A=-1\) and \(C+B=2\)

\(\therefore-1+B=0, C-2=-1\)

\(\Rightarrow B=1, C=1\)

\(\therefore \frac{2 x-1}{(x+1)\left(x^{2}+2\right)}=-\frac{1}{x+1}+\frac{x+1}{x^{2}+2}\)

Denominator with Repeated Irreducible Factors

Case IV : When \(g(x)\) contains repeated and non-repeated irreducible quadratic factors of the form \(\left(a x^{2}+b x+c\right)^{n}\)

Corresponding to every repeated irreducible quadratic factor of  \(g(x)\) there exist partial fractions of the form \(\frac{A_{1} x+B_{1}}{\left(a x^{2}+b x+c\right)}, \frac{A_{2} x+B_{2}}{\left(a x^{2}+b x+c\right)^{2}}, \cdots, \frac{A_{n} x+B_{n}}{\left(a x^{2}+b x+c\right)^{n}}\) where \(A_{1}, A_{2}, \ldots \ldots, A_{n}\), and \(B_{1}, B_{2}, \ldots, B_{n}\) are real numbers.

Example: Resolve \(\frac{2 x-3}{(x-1)\left(x^{2}+1\right)^{2}}\) into partial fractions.

Ans: Let

\(\frac{2 x-3}{(x-1)\left(x^{2}+1\right)^{2}}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+1}+\frac{D x+E}{\left(x^{2}+1\right)^{2}}\)

\(2 x-3=A\left(x^{2}+1\right)^{2}+(B x+C)(x-1)\left(x^{2}+1\right)+(D x+E)(x-1)\) ……(i)

\(= A\left( {{x^4} + 2{x^2} + 1} \right) + B{x^4} + ( – B + C){x^3} + (B – C){x^2} + (C – B)x – C + D{x^2} + Ex – Dx – E\)

\(=(A+B) x^{4}+(-B+C) x^{3}+(2 A+B-C+D) x^{2}+(C-B+E-D) x+A-C-E\) …..(ii)

Putting \(x=1\) in equation (i) we have,

\(-1=4 A\)

\(\Rightarrow A=-\frac{1}{4}\)

Equating coefficients of like powers of \(x\) from equation,we get 

\(A+B=0, C-B=0,2 A+B-C+D=0, C+E-B-D=2\) and \(A-C-E=-3\)

Putting \(A=-\frac{1}{4}\) and solving these equations, we get 

\(B=\frac{1}{4}=C, D=\frac{1}{2}\) and \(E=\frac{5}{2}\)

\(\therefore \frac{2 x-3}{(x-1)\left(x^{2}+1\right)^{2}}=\frac{-1}{4(x-1)}+\frac{x+1}{4\left(x^{2}+1\right)}+\frac{x+5}{2\left(x^{2}+1\right)^{2}}\)

Solved Examples on Integration by Partial Fractions

Q.1. Evaluate: \(\int \frac{x-1}{(x+1)(x-2)} d x\)
Ans: Let \(\frac{x-1}{(x+1)(x-2)}=\frac{A}{x+1}+\frac{B}{x-2}\) …..(i)
\(=\frac{A(x-2)+B(x+1)}{(x+1)(x-2)}\)
\(\Rightarrow x-1=A(x-2)+B(x+1)\)……(ii)
Substituting \(x=2\) in equation (ii), we have,
\(1=3 B\)
\(\Rightarrow B=\frac{1}{3}\)
Substituting \(x=-1\) in equation (ii), we have,
\(-2=-3 A\)
\(\therefore A=\frac{2}{3}\)
Thus, from (i), we have
\(\frac{x-1}{(x+1)(x-2)}=\frac{2}{3(x+1)}+\frac{1}{3(x-2)}\)
\(\therefore I=\int \frac{x-1}{(x+1)(x-2)} d x\)
\(=\int\left(\frac{2}{3(x+1)}+\frac{1}{3(x-2)}\right) d x\)
\(=\frac{2}{3} \int \frac{1}{x+1} d x+\frac{1}{3} \int \frac{1}{x-2} d x\)
\(\therefore \int \frac{x-1}{(x+1)(x-2)} d x=\frac{2}{3} \log |x+1|+\frac{1}{3} \log |x-2|+C\)

Q.2. Evaluate: \(\int \frac{3 x+1}{(x-2)^{2}(x+2)} d x\)
Ans: Let \(\frac{3 x+1}{(x-2)^{2}(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+2}\)…..(i)
\(=\frac{A(x-2)(x+2)+B(x+2)+C(x-2)^{2}}{(x-2)^{2}(x+2)}\)
\(\Rightarrow 3 x+1=A(x-2)(x+2)+B(x+2)+C(x-2)^{2}\) …….(ii)
Substituting \(x=2\) in equation (ii) we have,
\(7=4 B\)
\(\therefore B=\frac{7}{4}\)
Substituting \(x=-2\) in equation (ii), we have,
\(-5=16 C\)
\(\therefore C=-\frac{5}{16}\)
Putting \(x=0\) in equation (ii), we have
\(1=-4 A+2 B+4 C\)
\(\Rightarrow 4 A=2 B+4 C-1\)
\(=2 \times \frac{7}{4}+4 \times\left(-\frac{5}{16}\right)-1\)
\(=\frac{7}{2}-\frac{5}{4}-1\)
\(=\frac{5}{4}\)
\(4 A=\frac{5}{4}\)
\(\therefore A=\frac{5}{16}\)
Substituting the values of \(A,B\) and \(C\) in (i) we get,
\(\frac{3 x+1}{(x-2)^{2}(x+2)}=\frac{5}{16} \times \frac{1}{x-2}+\frac{7}{4} \frac{1}{(x-2)^{2}}-\frac{5}{16(x+2)}\)
\(\therefore I=\int \frac{3 x+1}{(x-2)^{2}(x+2)} d x\)
\(=\frac{5}{16} \int \frac{1}{x-2} d x+\frac{7}{4} \int \frac{1}{(x-2)^{2}} d x-\frac{5}{16} \int \frac{1}{x+2} d x\)
\(\therefore \int \frac{3 x+1}{(x-2)^{2}(x+2)} d x=\frac{5}{16} \log |x-2|-\frac{7}{4(x-2)}-\frac{5}{16} \log |x+2|+C\)

Q.3. Evaluate: \(\int \frac{8}{(x+2)\left(x^{2}+4\right)} d x\)
Ans: Let
\(\frac{8}{(x+2)\left(x^{2}+4\right)}=\frac{A}{x+2}+\frac{B x+C}{x^{2}+4}\) …..(i)
\(=\frac{A\left(x^{2}+4\right)+(B x+C)(x+2)}{(x+2)\left(x^{2}+4\right)}\)
\(\Rightarrow 8=A\left(x^{2}+4\right)+(B x+C)(x+2)\)…..(ii)
Substituting \(x=-2\) in equation (ii), we have
\(8=8 A\)
\(\Rightarrow A=1\)
Substituting \(x=0\) in euation (ii) we have
\(8=A(0+4)+2 C\)
\(\Rightarrow 4 A+2 C=8\)
\(\Rightarrow 2 C=8-4=4\)
\(\Rightarrow C=2\)
Substituting \(x=1\) in equation (ii) we have
\(8=A(1+4)+3(B+C)\)
\(\Rightarrow 8=5+3 B+6\) [Putting \(A=1, C=2\)]
\(\Rightarrow 3 B=8-11=-3\)
\(\Rightarrow B=-1\)
Thus, we have
\(A=1, C=2\) and \(B=-1\)
Substituting the values of \(A,B\) and \(C\) in (i), we obtain
\(\frac{8}{(x+2)\left(x^{2}+4\right)}=\frac{1}{x+2}+\frac{-x+2}{x^{2}+4}\)
\(\therefore I=\int \frac{8}{(x+2)\left(x^{2}+4\right)} d x\)
\(\Rightarrow I=\int \frac{1}{x+2} d x+\int \frac{-x+2}{x^{2}+4} d x\)
\(=\int \frac{1}{x+2} d x-\int \frac{x}{x^{2}+4} d x+2 \int \frac{1}{x^{2}+4} d x\)
\(=\log |x+2|-\frac{1}{2} \int \frac{1}{t} d t+2 \times \frac{1}{2} \tan ^{-1} \frac{x}{2}+C\), where \(t=x^{2}+4\)
\(=\log |x+2|-\frac{1}{2} \log t+\tan ^{-1} \frac{x}{2}+C\)
\(=\log |x+2|-\frac{1}{2} \log \left(x^{2}+4\right)+\tan ^{-1} \frac{x}{2}+C\)
Hence, \(\int \frac{8}{(x+2)\left(x^{2}+4\right)} d x=\log |x+2|-\frac{1}{2} \log \left(x^{2}+4\right)+\tan ^{-1} \frac{x}{2}+C\)

Q.4. Evaluate: \(\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x\)
Ans: Let \(x^{2}=y\), Then
\(\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{y}{(y+1)(y+4)}\)
Let
\(\frac{y}{(y+1)(y+4)}=\frac{A}{y+1}+\frac{B}{y+4}\)…..(i)
\(\Rightarrow y=A(y+4)+B(y+1)\)……(ii)
Substituting \(y=-1\) in equation (ii) we have
\(-1=3 A\)
\(\Rightarrow A=-\frac{1}{3}\)
Substituting \(y=-4\) in equation (ii) we have
\(-4=-3 B\)
\(\Rightarrow B=\frac{4}{3}\)
Substituting the values of \(A\) and \(B\) in equation (i), we have
\(\frac{y}{(y+1)(y+4)}=-\frac{1}{3(y+1)}+\frac{4}{3(y+4)}\)
Replacing \(y\) by \(x^{2}\), we obtain
\(\frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=-\frac{1}{3\left(x^{2}+1\right)}+\frac{4}{3\left(x^{2}+4\right)}\)
\(\therefore I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x\)
\( = \int {\left\{ { – \frac{1}{{3\left( {{x^2} + 1} \right)}} + \frac{4}{{3\left( {{x^2} + 4} \right)}}} \right\}} dx\)
\(=-\frac{1}{3} \int \frac{1}{x^{2}+1} d x+\frac{4}{3} \int \frac{1}{x^{2}+4} d x\)
\(=-\frac{1}{3} \tan ^{-1} x+\frac{4}{3} \times \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+C\)
\(\therefore \int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x=-\frac{1}{3} \tan ^{-1} x+\frac{2}{3} \tan ^{-1}\left(\frac{x}{2}\right)+C\)

Q.5. Evaluate: \(\int \frac{1}{x+\sqrt{x^{2}-x+1}} d x\)
Ans: Let \(I=\int \frac{1}{x+\sqrt{x^{2}-x+1}} d x\)…….(i)
Let \(x+\sqrt{x^{2}-x+1}=t\) Then
\(\sqrt{x^{2}-x+1}=t-x\)
\(\Rightarrow x^{2}-x+1=(t-x)^{2}\)
\(\Rightarrow-x+1=t^{2}-2 t x\)
\(\Rightarrow x=\frac{t^{2}-1}{2 t-1}\)
\(\therefore d x=\frac{(2 t-1) 2 t-2\left(t^{2}-1\right)}{(2 t-1)^{2}} d t\)
\(=\frac{2 t^{2}-2 t+2}{(2 t-1)^{2}} d t\)
Substituting the value of \(dx\) and \(x+\sqrt{x^{2}-x+1}\) in equation (i) we have,
\(I=\int\left[\frac{1}{t} \times \frac{2 t^{2}-2 t+2}{(2 t-1)^{2}}\right] d t\)
\(\therefore I=2 \int \frac{t^{2}-t+1}{t(2 t-1)^{2}} d t\)
Now, Let \(\frac{t^{2}-t+1}{t(2 t-1)^{2}}=\frac{A}{t}+\frac{B}{2 t-1}+\frac{C}{(2 t-1)^{2}}\) …..(ii)
\(\Rightarrow t^{2}-t+1=A(2 t-1)^{2}+B(2 t-1) t+C t\) ……(iii)
Substituting \(t=0\) in equation (iii), we have
\(1=A\)
Substituting \(t=\frac{1}{2}\) in equation (iii), we have
\(\frac{3}{4}=C \times \frac{1}{2}\)
\(\therefore C=\frac{3}{2}\)
Substituting \(t=1\) in equation (iii), we have,
\(1=A+B+C\)
\(\Rightarrow 1+B+\frac{3}{2}=1\)
\(\Rightarrow B+\frac{5}{2}=1\)
\(\therefore B=-\frac{3}{2}\)
Substituting the values of \(A, B\) and \(C\) in equation (ii), we have,
\(\frac{t^{2}-t+1}{t(2 t-1)^{2}}=\frac{1}{t}-\frac{3}{2(2 t-1)}+\frac{3}{2} \frac{1}{(2 t-1)^{2}}\)
\(\therefore I=2 \int \frac{1}{t} d t-3 \int \frac{1}{2 t-1} d t+3 \int \frac{1}{(2 t-1)^{2}} d t\)
\(=\log |t|-\frac{3}{2} \log |2 t-1|+\frac{3}{4} \times-\frac{1}{2 t-1}\)
\(\therefore \int \frac{1}{x+\sqrt{x^{2}-x+1}} d x\)
\(=\log \left|x+\sqrt{x^{2}-x+1}\right|-\frac{3}{2} \log \left|2\left(x+\sqrt{x^{2}-x+1}\right)-1\right|-\frac{3}{4} \times \frac{1}{2\left(x+\sqrt{x^{2}-x+1}\right)-1}\)

Summary

All proper rational functions \(\frac{f(x)}{g(x)}\) can be expressed as the sum of rational functions, each having a simple factor of \(g(x)\). Each such fraction is called a partial fraction and the process of obtaining them is called the resolution or decomposition of \(\frac{f(x)}{g(x)}\) into partial fractions. Resolution of \(\frac{f(x)}{g(x)}\) into partial fractions depends on the nature of the factors of \(g(x)\).

If the function is of the form \(\frac{p x+q}{(x-a)(x-b)}, a \neq b\) then we can write it as \(\frac{\mathrm{A}}{x-a}+\frac{\mathrm{B}}{x-b}\) where \(A\) and \(B\) are constants, whose values are need to be determined. Similarly, the rational function of the form \(\frac{p x+q}{(x-a)^{2}}\) can be written as \(\frac{\mathrm{A}}{x-a}+\frac{\mathrm{B}}{(x-a)^{2}}\)

Frequently Asked Questions (FAQs)

Q.1. What is the purpose of a partial fraction?
Ans: Sometimes the denominator in a rational fraction is very complex such that we cannot integrate it. So, the given expression can be written as the sum of simple fractions known as partial fractions. Thus, the essential notion behind partial fraction integration is to factor the denominator and then partition it into two or more fractions.

Q.2. Where are partial fractions used in real life?
Ans: Partial fractions are important in calculating the production, distribution, and consumption of commodities and services because they are used to break down complex rational equations into simpler ones. The change in the value of a function induced by a one-unit increase in one of its variables is determined by marginal analysis in economics.

Q.3. Why is partial fraction used in integration? 
Ans: We can only conduct partial fractions if the degree of the numerator is much smaller than that of the denominator. This is crucial. As a result, once you’ve grasped the concept of partial fractions, you will need to factor the denominator as much as possible.

Q.4. What is the rule of a partial fraction having a quadratic factor in the denominator?
Ans: Let \(\frac{f(x)}{g(x)}\) be a proper rational fraction such that
Case I: \(g(x)\) contains non-repeated irreducible factors of the form \(p x^{2}+q x+r\), then corresponding to every non-reducible non-repeated quadratic factor of \(g(x)\), there exists a partial fraction of the form \(\frac{A x+B}{p x^{2}+q x+r^{2}}\), where \(p, q, A\) and \(B\) are real numbers. 
Case II : \(g(x)\) contains repeated and non-repeated irreducible quadratic factors of the form \(\left(p x^{2}+q x+r\right)^{n}\) then  corresponding to every repeated irreducible quadratic factor of  \(g(x)\)
there  exist partial fractions of the form \(\frac{p_{1} x+q_{1}}{\left(p x^{2}+q x+r\right)}, \frac{p_{2} x+q_{2}}{\left(p x^{2}+q x+r\right)^{2}}, \cdots, \frac{p_{n} x+q_{n}}{\left(p x^{2}+q x+r\right)^{n}}\)
where \(p_{1}, p_{2}, \ldots, p_{n}\) and \(q_{1}, q_{2}, \ldots, q_{n}\) are real numbers.

Q.5. What is a partial fraction? Give example.
Ans: The fractions used to decompose a rational expression are called partial fractions. Each part of an algebraic expression is called a partial fraction when it is divided into a sum of two or more rational expressions.
Example: \(\frac{3 x+2}{(x-1)(x-2)(x-3)}=\frac{5}{2(x-1)}-\frac{8}{x-2}+\frac{11}{2(x-3)}\)
Here, each RHS term is called a partial fraction.

Q.6. How do you solve partial fractions with repeated roots?
Ans: Let \(\frac{f(x)}{g(x)}\) be a proper rational fraction such that \(g(x)\) contains some repeated linear factors.
Step 1: Let \(g(x)=(x-a)^{k}\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{r}\right)\)
Step 2: Assume that
\(\frac{f(x)}{g(x)}=\frac{A_{1}}{x-a}+\frac{A_{2}}{(x-a)^{2}}+\frac{A_{3}}{(x-a)^{3}}+\cdots+\frac{A_{k}}{(x-a)^{k}}+\frac{B_{1}}{x-a_{1}}+\frac{B_{2}}{x-a_{2}}+\cdots+\frac{B_{r}}{\left(x-a_{r}\right)}\)
Where \(A_{1}, A_{2}, \ldots, A_{k}, B_{1}, B_{2}, \ldots, B_{r}\) are constants.
Step 3: To determine constants, equate numerators on both sides. Some of the constants are determined by comparing coefficients of equal powers of \(x\) on both sides.

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