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Integration Using Partial Fraction - Practice Questions & MCQ

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

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Integration Using Partial Fraction is considered one the most difficult concept.
17 Questions around this concept.

Solve by difficulty

MEDIUM

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)

$\left ( \frac{8}{3} ,\frac{2}{3}\right )$

$\left ( -\frac{8}{3} ,\frac{2}{3}\right )$

$\left ( -\frac{8}{3},-\frac{2}{3} \right )$

$\left ( \frac{8}{3},-\frac{2}{3} \right )$

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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