Integration Using Partial Fraction - Practice Questions & MCQ

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

Partial fraction method is used to integrate rational functions.

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

Partial fraction decomposition can be applied to a rational function $\frac{P(x)}{Q(x)}$ only if deg $(\mathrm{P}(\mathrm{x}))<\operatorname{deg}(\mathrm{Q}(\mathrm{x}))$.
In the case when $\operatorname{deg}(\mathrm{P}(\mathrm{x})) \geq \operatorname{deg}(\mathrm{Q}(\mathrm{x}))$, 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 $\operatorname{deg}(\mathrm{P}(\mathrm{x})) \geq \operatorname{deg}(\mathrm{Q}(\mathrm{x}))$
$
\int \frac{\mathrm{x}^2+3 \mathrm{x}+5}{\mathrm{x}+1} \mathrm{dx}
$
since $\operatorname{deg}\left(x^2+3 x+5\right) \geq \operatorname{deg}(x+1)$, we perform long division to obtain
$
\begin{aligned}
& \frac{x^2+3 x+5}{x+1}=x+2+\frac{3}{x+1} \\
& \begin{aligned}
\int \frac{x^2+3 x+5}{x+1} d x & =\int\left(x+2+\frac{3}{x+1}\right) d x \\
& =\frac{1}{2} x^2+2 x+3 \ln |x+1|+C
\end{aligned}
\end{aligned}
$

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

Case 1: 

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

$\mathrm{Q}(\mathrm{x})$ can be factored as $\left(x-a_1\right)\left(x-a_2\right)\left(x-a_3\right) \ldots . . .\left(x-a_n\right)$ then, we have
$
\frac{P(x)}{Q(x)}=\frac{A_1}{\left(x-a_1\right)}+\frac{A_2}{\left(x-a_2\right)}+\cdots+\frac{A_n}{\left(x-a_n\right)}
$
where, $\mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots \ldots . \mathrm{A}_{\mathrm{n}}$ are constants and can be determined by equating the numerator on RHS to the numerator on LHS and then substituting
$
\mathrm{x}=\mathrm{a}_1, \mathrm{a}_2, \mathrm{a}_3 \ldots ., \mathrm{a}_{\mathrm{n}}
$ 

Short cut method

Consider $x-a_1=0$ then, $x=a_1$, put his value of x in all the expression other than $x-a_1$ and so on for example:
$
\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)}
$

Illustration 1: Evaluate $\int \frac{3 x+2}{x^3-x^2-2 x} d x$ since $\operatorname{deg}(3 x+2)<\operatorname{deg}\left(x^3-x^2-2 x\right)$, we begin by factoring the denominator of $\frac{3 x+2}{x^3-x^2-2 x}$.
$
\Rightarrow \quad \frac{3 x+2}{x(x-2)(x+1)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x+1}
$

We need to determine $A, B$, and $C$. Cross multiplying in the expression above, we obtain:
$
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 hold true for all x, we substitute those values of x that would straightway give us the required values of A, B, and C. These values are obviously the roots of Q(x).

$
\begin{aligned}
\mathrm{x}=0 & \Rightarrow 2=\mathrm{A}(0-2)(0+1)+\mathrm{B}(0)+\mathrm{C}(0) \\
& \Rightarrow \mathrm{A}=-1 \\
\mathrm{x}=2 & \Rightarrow 8=\mathrm{A}(0)+\mathrm{B}(2)(2+1)+\mathrm{C}(0) \\
& \Rightarrow \mathrm{B}=\frac{4}{3} \\
\mathrm{x}=-1 & \Rightarrow-1=\mathrm{A}(0)+\mathrm{B}(0)+\mathrm{C}(-1)(-1-2) \\
& \Rightarrow \mathrm{C}=-\frac{1}{3}
\end{aligned}
$

Now that we have the values of $\mathrm{A}, \mathrm{B}$, and C , we rewrite the original integral:
$
\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
$

Evaluating the integral gives us
$
\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 the denominator (Q(x)) is expressed as the product of the linear factor such that some of them are repeating (Linear and repeated)

$\mathrm{Q}(\mathrm{x})$ can be factored as $(x-a)^k\left(x-a_1\right)\left(x-a_2\right)\left(x-a_3\right) \ldots . .\left(x-a_n\right)$ then, we have
$
\frac{P(x)}{Q(x)}=\frac{A}{(x-a)}+\frac{A}{(x-a)^2}+\ldots.+\frac{A^k}{(x-a)^k}+\frac{B_1}{\left(x-a_1\right)}+\frac{B_2}{\left(x-a_2\right)}+\cdots+\frac{B_n}{\left(x-a_n\right)}
$

Case 3: 

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

Corresponding to each quadratic factor $a x^2+b x+c$, assume the 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 power of x in the numerator of both sides. or by substituting suitable values of x on both sides