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As we know, [itex]\int \frac {1}{x} \, dx = ln (x) + C[/itex]
But, what it would be, if [itex]x[/itex] is negative?
I calculated this way,
[itex]\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C[/itex] [because, [itex]-x[/itex] has a positive value.]
But, what it would be, if [itex]x[/itex] is negative?
I calculated this way,
[itex]\int \frac {1}{x} \, dx = - \int \frac {1}{-x} \, dx = \int \frac {1}{-x} \, d(-x) = ln (-x) + C[/itex] [because, [itex]-x[/itex] has a positive value.]