Evaluate the following intregals: 2x-3 (x^2-1)(2x-3) dx

Question

Evaluate the following intregals:
$\int\frac{2\text{x}-3}{(\text{x}^2-1)(2\text{x}-3)}\ \text{dx}$

✓

Answer

$\int\frac{2\text{x}-3}{(\text{x}^2-1)(2\text{x}-3)}\ \text{dx}$
$=\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}\ \text{dx}$
Let $\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}+1}+\frac{\text{C}}{2\text{x}+ 3}$
$\Rightarrow\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}=\frac{\text{A}(\text{x}+1)(2\text{x}+3)+\text{B}(\text{x}+1)(2\text {x}-3)+\text{C}(\text{x}^2-1)}{(\text{x}-1))(\text{x}+1)(2\text{X}+3)}$
$\Rightarrow2\text{x}-3=\text{A}(\text{x}+1)(2\text{x}+3)\\+\text{B}(\text{x}-1)(2\text{x}+3)+\text{C}(\text{x}+1)(\text{x}-1)\ ...(1)$
Putting x + 1 = 0 or x = -1 in eq (1)
⇒ -2 - 3 = B (-1 -1) (-2 + 3)
⇒ -5 = B (-2) (1)
$\Rightarrow\text{B}=\frac{5}{2}$
Putting x - 1 = 0 or x = 1 in eq (1)
⇒ 2 - 3 = A (1 + 1) (2 + 3)
⇒ -1 = A (2) (5)
$\Rightarrow\text{A}=\frac{-1}{10}$
Putting 2x + 3 = 0 or $\text{x}=\frac{-3}{2}$ in eq (1)
$\Rightarrow2\times-\frac{3}{2}-3=\text{A}\times0+\text{B}\times0+\text{C}\Big(-\frac{3}{2}+1\Big)\Big(\frac{-3}{2}-1\Big)$
$\Rightarrow-6=\text{C}\Big(-\frac{1}{2}\Big)\Big(\frac{-5}{2}\Big)$
$\Rightarrow\text{C}=-\frac{24}{5}$
$\therefore\frac{2\text{x}-3}{(\text{x}-1)(\text{X}+1)(2\text{X}+3)}=\frac{-1}{10(\text{x}-1)}+\frac{5}{2(\text{x}+1)}-\frac{24}{5(2\text{x}+3)}$
$\Rightarrow\int\frac{(2\text{x}-3)}{(\text{x}-1)(\text{x}+1)(2\text{x}+3)}\ \text{dx}=\frac{-1}{10}\int\frac{1}{\text{x}-1}\text{dx}\\+\frac{5}{2}\int\frac{1}{\text{x}+1}\text{dx}-\frac{24}{5}\int\frac{1}{2\text{X}+3}\ \text{dx}$
$=\frac{-1}{10}\ln|\text{x}-1|+\frac{5}{2}\ln|\text{x}+1|-\frac{24}{5}\ln\frac{|2\text{x}+3|}{3}+\text{C}$
$=-\frac{1}{10}\ln|\text{x}-1|+\frac{5}{2}\ln|\text{x}+1|-\frac{12}{5}\ln|2\text{x}+3|+\text{C}$