∫1/x dx = log|x| + C

The power formula for $x^n$ fails at $n=-1$ because the denominator becomes $0$. $\ln|x|$ fills that gap.

The absolute value is there so the formula also works for negative $x$.

Examples

1. 01For which n does this formula fail?$\displaystyle \int x^{n}\,dx = \frac{x^{n+1}}{n+1}$AnswerHide

   Answer$\displaystyle n=-1$

   Steps

   1. Find the $n$ that makes the denominator $n+1$ zero
   2. $\displaystyle n+1 = 0$$\displaystyle \Rightarrow n = -1$
   3. For every other $n$ the denominator is not 0, so the rule works as written
2. 02What is this?$\displaystyle \int \frac{2}{x}\,dx$AnswerHide

   Answer$\displaystyle 2\log_e|x|+C$

   Steps

   1. A constant factor just comes along; $\dfrac{1}{x}$ integrates to $\log_e|x|$
   2. $\displaystyle f(x)$$\displaystyle = 2\log_e|x|$$\displaystyle f'(x)$$\displaystyle = \frac{2}{x}$
   3. Differentiating gets us back. The $2$ stays in front of the $\log$, never inside it
3. 03What is the value?$\displaystyle \int_{1}^{e} \frac{4}{x}\,dx$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. $\dfrac{1}{x}$ integrates to $\log_e|x|$; subtract the lower end from the upper
   2. $\displaystyle 4(\log_e e-\log_e 1)$$\displaystyle = 4(1-0)$$\displaystyle = 4$
   3. $\log_e e = 1$ and $\log_e 1 = 0$; taking $e$ as the upper end makes the answer a whole number
4. 04What goes in □?$\displaystyle \int_{1}^{3} \frac{3}{x}\,dx = \square\log_e 3$AnswerHide

   Answer$\displaystyle 3$

   Steps

   1. The constant comes out and $\dfrac{1}{x}$ integrates to $\log_e|x|$, so subtract the lower end from the upper
   2. $\displaystyle 3(\log_e 3-\log_e 1)$$\displaystyle = 3\log_e 3$
   3. $\log_e 1 = 0$, so only $\log_e 3$ survives; $\square$ is $3$
5. 05Integrate$\displaystyle \int \frac{1}{x} \, dx$AnswerHide

   Answer$\displaystyle \log_e|x| + C$
6. 06Find the value$\displaystyle \int_{1}^{4} \frac{1}{x^2} \, dx$AnswerHide

   Answer$\displaystyle \frac{3}{4}$

FAQ