Indefinite Integrals — Maths STD 12 Science — Question

$\text{I}=\int\frac{\text{x}\ \text{dx}}{(\text{x}^2-\text{a}^2)(\text{x}^2-\text{b}^2)}$

Putting x² = t

$⇒ 2\text{xdx} = \text{dt}$

$\Rightarrow\text{xdx}=\frac{\text{dt}}{2}$

$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$

Let $\frac{1}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}=\frac{\text{A}}{\text{t}-\text{a}^2}+\frac{\text{B}}{\text{t}-\text{b}^2}$

$\Rightarrow\frac{1}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}=\frac{\text{A}(\text{t}-\text{b}^2)+\text{B}(\text{t}-\text{a}^2)}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$

$\Rightarrow1=\text{A}(\text{t}-\text{b}^2)+\text{B}(\text{t}-\text{a}^2)$

Putting t = b²

$1=\text{A}(\text{a}^2-\text{b}^2)+\text{B}\times0$

$\Rightarrow\text{A}=\frac{1}{\text{a}^2-\text{b}^2}$

$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$

$=\frac{1}{2(\text{a}^2-\text{b}^2)}\int\frac{}{}\frac{\text{dt}}{\text{t}-\text{a}^2}+\frac{1}{2((\text{b}^2-\text{a}^2)}\int\frac{\text{dt}}{\text{t}-\text{b}^2}$

$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|+\frac{1}{2(\text{b}^2-\text{a}^2)}\log|\text{t}-\text{b}^2|+\text{C}$

$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|-\log|\text{t}-\text{b}^2|]+\text{C}$

$=\frac{1}{2(\text{a}^2-\text{b}^2)}\Big[\log\Big|\frac{\text{t}-\text{a}^2}{\text{t}-\text{b}^2}\Big|\Big]+\text{C}$

$=\frac{1}{2(\text{a}^2-\text{b}^2)}\log\Big|\frac{\text{x}^2-\text{a}^2}{\text{x}^2-\text{b}^2}\Big|+\text{C}$