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-- Mathematics - Section A
Indefinite Integration
Find the integral \(\int\frac{1}{3x}dx\)
\[\frac{1}{ { 3x }^{ 2 } } + c\]
\[\frac{1}{ { 3x }^{ -2 } } + c\]
3x +c
\[\frac{log x}{ { 3 } } + c\]
\(\int\frac{1}{ax}dx\) = \(\frac{log x}{ { a } } + c\)