MCQ
Choose the correct answer in exercise:
$\int\frac{\text{x dx}}{(\text{x}-1)(\text{x}-2)}$ equals
  • A
    $\text{log}\Bigg|\frac{(\text{x}-1)^2}{\text{x}-2}\Bigg|+\text{C}$
  • $\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$
  • C
    $\text{log}\Bigg|\Bigg(\frac{\text{x}-1}{\text{x}-2}\Bigg)^2\Bigg|+\text{C}$
  • D
    $\text{log}|(\text{x}-1)(\text{x}-2)|+\text{c}$

Answer

Correct option: B.
$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$
$\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{C}$Let, $\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}\dots(\text{i})$
⇒ x = A(x - 2) + B(x -1)
⇒ x = Ax - 2A + Bx - B
Comparing coefficients of x:
A + B = 1 ……….(ii)
Comparing constants
–2A – B = 0 ……….(iii)
On solving eq. (ii) and (iii), we get
A = –1, B = 2
Putting these values of A and B in eq. (i),
$\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}=\frac{-1}{\text{x}-1}+\frac{2}{\text{x}-2}$
$\Rightarrow\ \int\frac{\text{x}}{(\text{x}-1)(\text{x}-2)}\text{dx}=-\int\frac{1}{\text{x}-1}\text{dx}+2\int\frac{1}{\text{x}-2}\text{dx}$
$=-\text{log}|\text{x}-1|+2\text{log}|\text{x}-2|+\text{c}$
$=\text{log}|(\text{x}-2)^2|-\text{log}|\text{x}-1|+\text{c}=\text{log}\Bigg|\frac{(\text{x}-2)^2}{\text{x}-1}\Bigg|+\text{c}$

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