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### Integrating x^-1 or 1/x

You may be tempted to apply the power rule of integration to x^-1, but you can't.

This can be understood by trying to apply the power rule to x^-1 and seeing what you get. If you apply the power rule to x^-1, on integrating you get division by zero:
\int x^-1 dx = x^(-1 + 1) / (-1 + 1) = x^0 / 0 = 1/0
Division by zero does not give you a well defined answer, so this method is not appropriate for integrating 1/x. Instead there is a formula, as follows:
\int 1/x dx = ln|x| + C
... where ln|x| is the natural logarithmic function. It is same as log_e|x|. Note that ln|x| has absolute value bars around 'x' in the above formula. Examples usages of this formula are:

1. \int 1/(2x) dx = ln|2x| /2 + C
2. \int 1/(3x + 7) dx = ln|3x + 7| /3 + C
3. \int 1/(x - 1) dx = ln|x - 1| + C

Note: Simple substitution is used in the above example, which is discussed in the next post.

Next Post: Simple substitution in integrals