Partial fraction of integral giving no solution.

Question

The below integral is to be integrated using partial fraction decomposition $$\int \frac{3x + 5}{(x-1)^2(x+1)} \, dx$$ I tried to form a partial fraction in the following manner and ended up getting a system of two equations with no solution. $$ \frac{3x+5}{(x-1)^2(x+1)} = \frac{A}{(x-1)^2} + \frac{B}{x+1}$$ On checking the solution given in my school textbook I found out that the correct expression for partial fraction is: $$ \frac{3x+5}{(x-1)^2(x+1)} = \frac{A}{(x-1)^2} + \frac{B}{x+1} + \frac{C}{x-1}$$ I don't understand why the first partial fraction expression doesn't give me any solutions while the second does. I would like to know what the difference between the two partial fraction expressions are and why is it that the second way works fine.

Answer 1

Let us try to understand the situation by using example.

If we add up $\frac{1}{x-1}$, $\frac{1}{(x-1)^2}$ and $\frac{1}{x+1}$, we will get $\frac{2x^2-2x+1}{(x-1)^2(x+1)}$.

That is

$$\frac{1}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{x+1}=\frac{2x^2-2x+1}{(x-1)^2(x+1)}.$$

The task of partial fraction is to start from $$\frac{2x^2-2x+1}{(x-1)^2(x+1)}$$ and try to get back

$$\frac{1}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{x+1}$$

Obviously we can never come to $\frac{1}{x-1}+\frac{1}{(x-1)^2}+\frac{1}{x+1}$ by assuming that the answer is of the form $\frac{A}{(x-1)^2}+\frac{B}{x+1}.$

The whole idea is to let $$\frac{2x^2-2x+1}{(x-1)^2(x+1)}=\frac{P(x)}{(x-1)^2}+\frac{Q(x)}{x+1}$$ where $\deg P(x) \lt \deg (x-1)^2$ and $\deg Q(x) \lt \deg (x+1).$

Thus $P(x)=Ax+B$ and $Q(x)=C$ for some constants $A, B$ and $C$.

Hence we have $$\frac{2x^2-2x+1}{(x-1)^2(x+1)}=\frac{Ax+B}{(x-1)^2}+\frac{C}{x+1}.$$

Since $Ax+B=A(x-1)+(A+B)$, we have $$\frac{2x^2-2x+1}{(x-1)^2(x+1)}=\frac{A(x-1)+A+B}{(x-1)^2}+\frac{C}{x+1}$$ i.e. $$\frac{2x^2-2x+1}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B'}{(x-1)^2}+\frac{C}{x+1}$$ where $B'=A+B.$

Answer 2

The first attempt produces $3$ simultaneous equations with $2$ unknowns, but this leads to inconsistency.

$\qquad{\begin{align}\dfrac{3x+5}{(x+1)(x-1)^2} &= \dfrac{A}{(x-1)^2} + \dfrac{B}{x+1} \\&=\dfrac{A(x+1)+B(x-1)^2}{(x+1)(x-1)^2}\\&=\dfrac{Bx^2+(A-2B)x+(A+B)}{(x+1)(x-1)^2}\end{align}\\\therefore\begin{cases}0 &= B\\3 &=A-2B\\5&= A+B\end{cases}\\\text{But $3\neq 5$}}$

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To obtain a solution, we would need $3$ unknowns in $3$ linear equations. That count of equations is because the denominator is a cubic expression, and so too is the numerator (albeit with a coefficient of $0$ for the $x^2$ term).

So how do we add a third unknown? Just add a third quotient whose determinant is also a factor of the original determinant.

$\qquad{\begin{align}\dfrac{3x+5}{(x+1)(x-1)^2} &= \dfrac{A}{(x-1)^2} + \dfrac{B}{x+1} +\dfrac{C}{(x-1)} \\&=\dfrac{A(x+1)+B(x-1)^2+C(x+1)(x-1)}{(x+1)(x-1)^2}\\&=\dfrac{(B+C)x^2+(A-2B)x+(A+B-C)}{(x+1)(x-1)^2}\end{align}\\\therefore\begin{cases}0 &= B+C\\3 &=A-2B\\5&= A+B-C\end{cases}}$

Hence $A=4$, $B=1/2$, and $C=-1/2$.

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Alternatively, we could increase the degree for the numerator of the first term.

$\qquad{\begin{align}\dfrac{3x+5}{(x+1)(x-1)^2} &= \dfrac{E x+F}{(x-1)^2} + \dfrac{B}{x+1} \\&=\dfrac{(E x+F)(x+1)+B(x-1)^2}{(x+1)(x-1)^2}\\&=\dfrac{(E+B)x^2+(E+F-2B)x+(F+B)}{(x+1)(x-1)^2}\end{align}\\\therefore\begin{cases}0 &= E+B\\3 &=E+F-2B\\5&= F+B\end{cases}}$

So $E = -1/2$, $F=9/2$, and $B = 1/2$

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Hence $E=C$ and $F=A-C$ , as should be anticipated since $\tfrac {Cx+A-C}{(x-1)^2}=\tfrac{A}{(x-1)^2}+\tfrac{C}{x-1}$