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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{\text{x}}{(\text{x}^2-\text{a}^2)(\text{x}^2-\text{b}^2)}\ \text{dx}$
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Answer

We have,$\text{I}=\int\frac{\text{x}\ \text{dx}}{(\text{x}^2-\text{a}^2)(\text{x}^2-\text{b}^2)}$
Putting $x^2= t$
$⇒ 2\text{xdx} = \text{dt}$
$\Rightarrow\text{xdx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$
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 $t = b^2$​​​​​​​
$1=\text{A}(\text{a}^2-\text{b}^2)+\text{B}\times0$
$\Rightarrow\text{A}=\frac{1}{\text{a}^2-\text{b}^2}$
$\text{I}=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}-\text{a}^2)(\text{t}-\text{b}^2)}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\int\frac{}{}\frac{\text{dt}}{\text{t}-\text{a}^2}+\frac{1}{2((\text{b}^2-\text{a}^2)}\int\frac{\text{dt}}{\text{t}-\text{b}^2}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|+\frac{1}{2(\text{b}^2-\text{a}^2)}\log|\text{t}-\text{b}^2|+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}[\log|\text{t}-\text{a}^2|-\log|\text{t}-\text{b}^2|]+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\Big[\log\Big|\frac{\text{t}-\text{a}^2}{\text{t}-\text{b}^2}\Big|\Big]+\text{C}$
$=\frac{1}{2(\text{a}^2-\text{b}^2)}\log\Big|\frac{\text{x}^2-\text{a}^2}{\text{x}^2-\text{b}^2}\Big|+\text{C}$

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