Careers360 Logo
JCECE 2025 - Date, Registration, Seat Allotment, Choice Filling, Merit List, Counselling, Cutoff

Integration Using Partial Fraction - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Integration Using Partial Fraction is considered one the most difficult concept.

  • 22 Questions around this concept.

Solve by difficulty

 If \small f\left ( \frac{3x-4}{3x+4} \right ) =x+2,\: \: x\neq -\frac{4}{3} and \small \int f\left ( x \right )dx =A \log \left | 1-x \right |+Bx+C then the ordered pair (A, B) is equal to : (where C is a constant of integration)

Evaluate  \int\frac{4x+1}{x^{2}-36}dx

Let f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x.If f(3)=\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right), then f(4) is equal to

Concepts Covered - 1

Integration Using Partial Fraction

A rational function is of the form P(x) / Q(x), where P(x) and Q(x) are polynomial function and Q(x) ≠ 0.

Partial fraction method is used to integrate rational functions.

\\\mathrm{Expression\;\mathit{\frac{P(x)}{Q(x)}} \;can \;be\;decomposed\;into\;simpler\;rational\;expression\;that\;we\;can}\\\mathrm{add\;or\;subtract\;to\;get\;the\;original\;rational\;function.\;This\;process\;is\;called}\\\mathrm{partial\;fraction\;decomposition.}

Partial fraction decomposition can be applied to a rational function \frac{P(x)}{Q(x)} only if deg (P(x)) < deg (Q(x)).

In the case when deg (P(x)) ≥ deg(Q(x)), we must first perform long division to rewrite the quotient \frac{P(x)}{Q(x)} in the form of A(x)+\frac{R(x)}{Q(x)}. where deg (R(x)) < deg (Q(x)). We then do a partial fraction decomposition on \frac{R(x)}{Q(x)}.

 

Illustration on how to approach to integrals of rational functions of the form \int \frac{P(x)}{Q(x)} d x where, deg (P(x)) ≥ deg(Q(x))

\\\mathrm{\int \frac{x^{2}+3 x+5}{x+1} d x}\\\text { since } \operatorname{deg}\left(x^{2}+3 x+5\right) \geq \operatorname{deg}(x+1), \text { we perform long division to obtain }\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{x^{2}+3 x+5}{x+1}=x+2+\frac{3}{x+1}}\\\text{thus,}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int \frac{x^{2}+3 x+5}{x+1} d x=\int\left(x+2+\frac{3}{x+1}\right) d x}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{2} x^{2}+2 x+3 \ln |x+1|+C}

 

If the rational function \frac{P(x)}{Q(x)} is proper (i.e. deg (P(x)) < deg (Q(x)) ), then three cases arises

 

Case 1: 

When denominator (Q(x)) is expressed as the product of the non-repeated linear factor

\\\mathrm{Q(x) \text { can be factored as }} (x-a_1)(x-a_2)(x-a_3).....(x-a_n) \\\text{then, we have}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\frac{P(x)}{Q(x)}=\frac{A_{1}}{(x-a_1)}+\frac{A_{2}}{(x-a_2)}+\cdots+\frac{A_{n}}{(x-a_n)}}\\\\\mathrm{where,\;A_1\;,A_2,\;A_3,......,\;A_n\;\text{are constants and can be determined by equating }}\\\text{the numerator on RHS to numerator on LHS and then substituting}\\\mathrm{x=a_1,\;a_2,\;a_3....,\;a_n} 

 

Short cut method

\\\text{Consider }x-a_1=0\text{ then, }x=a_1,\;\text{put his value of x in all the }\\\text{expression other than }x-a_1\;\text{and so on}\\\mathrm{for\;example:}\\\\\mathrm{\frac{x^2+2}{x(x-1)(x-2)}=\frac{0+2}{x(0-1)(0-2)}+\frac{1+2}{1(x-1)(1-2)}+\frac{4+2}{2(2-1)(x-2)}}


 

\\\text{Illustration 1: Evaluate} \int \frac{3 x+2}{x^{3}-x^{2}-2 x} d x\\\\\text { since } \operatorname{deg}(3 x+2)<\operatorname{deg}\left(x^{3}-x^{2}-2 x\right), \text { we begin by factoring the denominator of }\\\frac{3 x+2}{x^{3}-x^{2}-2 x}.\\\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{3 x+2}{x(x-2)(x+1)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+1}\\ {\text { We need to determine } A, B \text { and } C . \text { Cross multiplying in the expression above, }} \\ {\text {we obtain: }}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}3 x+2=A(x-2)(x+1)+B x(x+1)+C x(x-2)

A, B, C  can now be determined by

Option 1 : comparing coefficients of different powers of x and the constant terms on both sides.

Option 2: Since the relation that we just obtained should held true for all x, we substitute those values of x that would straight way give us the required values of A, B and C. These values are obviously the roots of Q(x).

\\\mathrm{x=0\;\;\;\;\;\;\Rightarrow \;\;\;\;2=A(0-2)(0+1)+B(0)+C(0)}\\\mathrm{\quad\quad\;\;\;\;\;\;\;\;\Rightarrow \;\;\;\;A=-1}\\\\\mathrm{x=2\;\;\;\;\;\;\Rightarrow \;\;\;\;8=A(0)+B(2)(2+1)+C(0)}\\\mathrm{\quad\quad\;\;\;\;\;\;\;\;\Rightarrow \;\;\;\;B=\frac{4}{3}}\\\\\mathrm{x=-1\;\;\;\;\;\;\Rightarrow \;\;\;\;-1=A(0)+B(0)+C(-1)(-1-2)}\\\mathrm{\quad\quad\;\;\;\;\;\;\;\;\Rightarrow \;\;\;\;C=-\frac{1}{3}}\\\text{Now that we have the values of A, B, and C, we rewrite the original integral:}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}\int \frac{3 x+2}{x^{3}-x^{2}-2 x} d x=\int\left(-\frac{1}{x}+\frac{4}{3} \cdot \frac{1}{(x-2)}-\frac{1}{3} \cdot \frac{1}{(x+1)}\right) d x\\\\\text{Evaluating the integral gives us}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}\int \frac{3 x+2}{x^{3}-x^{2}-2 x} d x=-\ln |x|+\frac{4}{3} \ln |x-2|-\frac{1}{3} \ln |x+1|+C

 

Case 2: 

When denominator (Q(x)) is expressed as the product of the linear factor such that some of them are repeating (Linear and repeated)

\\\mathrm{Q(x) \text { can be factored as }} (x-a)^k(x-a_1)(x-a_2)(x-a_3).....(x-a_n) \\\text{then, we have}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\frac{P(x)}{Q(x)}=\frac{A}{(x-a)}+\frac{A}{(x-a)^2}+....+\frac{A^k}{(x-a)^k}+\frac{B_{1}}{(x-a_1)}+\frac{B_{2}}{(x-a_2)}+\cdots+\frac{B_{n}}{(x-a_n)}}

 

Case 3: 

When some of the factors in denominator (Q(x)) are quadratic but non-repeating 

Corresponding to each quadratic factor ax2 + bx + c, assume the partial fraction of the type  

\frac{Ax+B}{ax^2+bx+c} , where A and B are constants to be determined by comparing coefficients of similar power of x in the numerator of  both sides. or by substituting suitable values of x on both sides

Study it with Videos

Integration Using Partial Fraction

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Books

Reference Books

Integration Using Partial Fraction

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.22

Line : 31

E-books & Sample Papers

Get Answer to all your questions

Back to top