how to resolve this integral using partial fractions

Question

I cannot solve this, please help me and explain me! $$\int \frac {2x-3} {x(2x+1)(2x-1)}\text{d}x$$

EDIT: I get this far

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

Answer 1

Let us set $f(z)=\dfrac{2z-3}{z(2z+1)(2z-1)}$. Then: $$\text{Res}\left(f(z),z=-\frac{1}{2}\right)=-2,$$ $$\text{Res}\left(f(z),z=0\right)=3,$$ $$\text{Res}\left(f(z),z=\frac{1}{2}\right)=-1,$$ hence: $$ f(z) = -\frac{2}{z+\frac{1}{2}}+\frac{3}{z}-\frac{1}{z-\frac{1}{2}}$$ so: $$ \int f(z)\,dz = C -2\log(z+1/2)+3\log z-\log(z-1/2).$$

Answer 2

$$ \frac {2x-3} {x(2x+1)(2x-1)} = \frac{A}{x} + \frac{B}{2x+1}+\frac{C}{2x-1} $$ Let us add the fractions, using a common denominator: \begin{align} & \frac{A(2x+1)(2x-1) + Bx(2x-1) + Cx(2x+1)}{x(2x+1)(2x-1)} \\[10pt] = {} & \frac{(4A+2B+2C)x^2 + (-B+C)x + (-A-B+C)}{x(2x+1)(2x-1)} \end{align} The numerator, $(4A+2B+2C)x^2 + (-B+C)x + (-A-B+C)$, should be equal to the original numerator $0x^2+2x+(-3)$. Hence we should have \begin{align} 4A+2B+2C & = 0 \\ -B+C & = 2 \\ -A-B+C & = -3 \end{align} Can you do the rest?

Answer 3

Here is a quick way to evaluate the coefficients of the partial fraction expansion. First we write

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

Now, we find $A$ as follows. Multiply both sides of $(1)$ by $x$ and take a limit of both resulting sides as $x \to 0$. The left side is

$$\lim_{x\to 0}\,x\,\frac{2x-3}{x(2x+1)(2x-1)}=\frac{-3}{-1}=3$$

and the right side is

$$\lim_{x\to 0}\,x\,\left(\frac{A}{x}+\frac{B}{2x+1}+\frac{C}{2x-1}\right)=A$$

Thus, $A=3$

---

Now, we find $B$ similarly. Multiply both sides of $(1)$ by $2x+1$ and take a limit of both resulting sides as $x \to -\frac12$. The left side is

$$\lim_{x\to -1/2}(2x+1)\frac{2x-3}{x(2x+1)(2x-1)}=\frac{-4}{1}=-4$$

and the right side is

$$\lim_{x\to -1/2}\,(2x+1)\,\left(\frac{A}{x}+\frac{B}{2x+1}+\frac{C}{2x-1}\right)=B$$

Thus, $B=-4$

---

Finally, we find $C$ using the same approach. Multiply both sides of $(1)$ by $2x-1$ and take a limit of both resulting sides as $x \to \frac12$. The left side is

$$\lim_{x\to 1/2}(2x-1)\frac{2x-3}{x(2x+1)(2x-1)}=\frac{-4}{1}=-2$$

and the right side is

$$\lim_{x\to 1/2}\,(2x-1)\,\left(\frac{A}{x}+\frac{B}{2x+1}+\frac{C}{2x-1}\right)=C$$

Thus, $C=-2$

---

NOTE: This approach is effectively the same as calculating residues. Yet, it relies on a perhaps more elementary exposition.