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:

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`

Next Post: Simple substitution in integrals

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`

**Simple substitution is used in the above example, which is discussed in the next post.**__Note__:Next Post: Simple substitution in integrals