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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
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Answer

$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Let $\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-3}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2-1)(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-2)(\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)+\text{C}(\text{x}-1)(\text{x}-2)}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}-2) (\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)\\+\text{C}(\text{x}-1)(\text{x}-2)\ \dots(1)$
Putting x - 1 = 0 or x = 1 in eq (1)
⇒ 1 = A (1 - 2) (1 - 3)
⇒ 1 = A (-1) (-2)
$\text{A}=\frac{1}{2}$
Putting x - 2 = 0 or x = 2 in eq (1)
⇒ 4 = B (2 - 1)(2 - 3)
⇒ B = -4
Putting x - 3 = 0 or x = 3 in eq (1)
⇒ 9 = C (3 - 1) (3 - 2)
$\Rightarrow\text{C}=\frac{9}{2}$
$\therefore\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}=\frac{1}{2(\text{x}-1)}-\frac{4}{\text{x}-2}+\frac{9}{2(\text{x}-3)}$
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{1}{2}\int\frac{1}{\text{x}-1}\ \text{dx}-4\int\frac{1}{\text{x}-2}+\frac{9}{2(\text{x}-3)}\ \text{dx}$
$=\frac{1}{2}\ln|\text{x}-1|-4\ln|\text{x}-2|+\frac{9}{2}\ln|\text{x}-3|+\text{C}$

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