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CUET
-- Mathematics - Section A
Indefinite Integration
Find the integral \( \int \frac{ 1}{ x} dx\)
\[\frac{1}{ { x }^{ 2 } } + c\]
\[\frac{1}{ { x }^{ -2 } } + c\]
x + c
log x + c
\( \int \frac{ 1}{ x} dx\) = log x +c