Find the indefinite integral of 3/x - 2/x^2 with respect to x.

Step-by-step solution

1. 1

   Split by linearity and pull the constants out.

   $\int\left(\frac{3}{x}-\frac{2}{x^2}\right)dx=3\int\frac{dx}{x}-2\int\frac{dx}{x^2}$

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

2. 2

   First term: the one case the power rule cannot handle. Writing $\tfrac1x=x^{-1}$, the rule $\int x^ndx=\tfrac{x^{n+1}}{n+1}$ would divide by $n+1=0$. So $n=-1$ is the exception, handled by the logarithm:

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

   The absolute value is what keeps this valid for negative $x$. The first term therefore contributes $3\ln|x|$.

3. 3

   Second term: the power rule applies normally. Here $\tfrac{1}{x^2}=x^{-2}$ and $n+1=-1\neq0$:

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

   so $2\int x^{-2}dx=-\dfrac{2}{x}$.

4. 4

   Apply the leading minus sign. The integrand subtracts that second piece:

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

   The sign flip is the step most often dropped — the answer has $+\tfrac2x$, not $-\tfrac2x$.

5. 5

   Combine into a single antiderivative.

   $\int\left(\frac{3}{x}-\frac{2}{x^2}\right)dx=3\ln|x|+\frac{2}{x}+C$

   Two similar-looking terms produce two entirely different function types — a logarithm and a rational term.

6. 6

   Verify by differentiating. $\dfrac{d}{dx}\left[3\ln|x|+\tfrac2x\right]=\dfrac{3}{x}-\dfrac{2}{x^2}$ ✓. Numerically, symmetric difference quotients at $x=0.7,\ 2.3,\ -1.6$ match the integrand to five digits ✓, including a negative $x$.