MCQ
$\int(1-x)^{-2} d x=$
  • $(1-x)^{-1}+c$
  • B
    $(1+x)^{-1}+c$
  • C
    $(1-x)^{-1}-1+c$
  • D
    $(1-x)^{-1}+1+c$

Answer

Correct option: A.
$(1-x)^{-1}+c$
$\int(1-x)^{-2} d x=\underline{( 1 - x )^{- 1 }+ c .}$
Explanation:
$\int(1-x)^{-2} d x=\frac{(1-x)^{-2+1}}{(-2+1)} \cdot \frac{d}{d x}(1-x)+c$
$=\frac{(1-x)^{-1}}{(-1)} \cdot(0-1)+c \$
$=(1- x )^{-1}+c$

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