Find the indefinite integral of 2/x - 1/x^2 with respect to x.

Step-by-step solution

1. 1

   Split the integral using linearity. Integration distributes over a difference, and constants come out front:

   $\int\left(\frac{2}{x}-\frac{1}{x^2}\right)dx=2\int\frac{dx}{x}-\int\frac{dx}{x^2}$

   The domain is $x\neq0$, and the antiderivative is valid separately on $(-\infty,0)$ and $(0,\infty)$.

2. 2

   Handle the first term — the power rule exception. Writing $\tfrac1x=x^{-1}$, the usual formula $\int x^n dx=\tfrac{x^{n+1}}{n+1}$ would divide by $n+1=0$. That is why $n=-1$ is the single exception:

   $\int\frac{dx}{x}=\ln|x|+C$

   The absolute value keeps it valid for negative $x$ too. So the first term contributes $2\ln|x|$.

3. 3

   Handle the second term with the ordinary power rule. Here $\tfrac{1}{x^2}=x^{-2}$ and $n=-2$, so $n+1=-1\neq0$ and the standard rule applies:

   $\int x^{-2}dx=\frac{x^{-1}}{-1}=-\frac1x$

4. 4

   Apply the minus sign in front of the second integral. The integrand has $-\tfrac{1}{x^2}$, so

   $-\int x^{-2}dx=-\left(-\frac1x\right)=+\frac1x$

   The double negative turning into a plus is the step most often lost.

5. 5

   Combine into one antiderivative.

   $\int\left(\frac{2}{x}-\frac{1}{x^2}\right)dx=2\ln|x|+\frac1x+C$

   Note how two superficially similar terms produce completely different function types — a logarithm and a rational term.

6. 6

   Verify by differentiating. $\dfrac{d}{dx}\left[2\ln|x|+\tfrac1x\right]=\dfrac{2}{x}-\dfrac{1}{x^2}$ ✓. Numerically, symmetric difference quotients at $x=0.7,\ 2.3,\ -1.6$ match the integrand to five digits ✓, including a negative $x$ where the absolute value matters.