Find the integral of 1/x with respect to x.

Step-by-step solution

1. 1

   See why the power rule fails here. The rule $\int x^{p}dx=\dfrac{x^{p+1}}{p+1}$ breaks down at $p=-1$, because the denominator $p+1$ becomes zero. So $\int x^{-1}dx$ is the one exception in the whole family and needs its own answer.

2. 2

   Start from the derivative of the logarithm. For $x>0$, $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$, so $\ln x$ is an antiderivative — on the positive axis only, since $\ln x$ is undefined for $x\le 0$.

3. 3

   Handle negative $x$, where the integrand is still fine. $\dfrac1x$ is perfectly well defined for $x<0$, so it must have an antiderivative there too. For $x<0$ write $|x|=-x$ and differentiate with the chain rule:

   $\frac{d}{dx}\ln(-x)=\frac{1}{-x}\cdot(-1)=\frac{1}{x}$

   The same derivative comes out, which is exactly why the absolute value is the right patch.

4. 4

   Combine both cases into one formula.

   $\int\frac{1}{x}\,dx=\ln\left|x\right|+C\qquad(x\neq 0)$

   Writing $\ln x+C$ instead would silently restrict the answer to $x>0$ and give an undefined expression for every negative input.

5. 5

   Note a subtlety about the constant. Because the domain $x\neq 0$ is split into two disconnected pieces, the constant may in fact differ on each side: the fully general antiderivative is $\ln|x|+C_{1}$ for $x>0$ and $\ln|x|+C_{2}$ for $x<0$. For definite integrals this matters — an integral crossing $0$, such as $\int_{-1}^{1}\frac{dx}{x}$, is improper and divergent, not $\ln 1-\ln 1=0$.