Evaluate the integral of 2 dy divided by -2y.

Step-by-step solution

1. 1

   Cancel the numerical constants first. The $2$ on top and the $-2$ underneath reduce to $-1$:

   $\int \frac{2}{-2y}\,dy = \int \left(-\frac{1}{y}\right)dy$

   Doing this before integrating avoids carrying factors through the antiderivative.

2. 2

   Pull the constant out. Constants factor through an integral:

   $\int \left(-\frac{1}{y}\right)dy = -\int \frac{1}{y}\,dy$

3. 3

   Apply the reciprocal rule.

   $\int \frac{1}{y}\,dy = \ln|y| + C$

   The absolute value matters: $\tfrac1y$ is defined for negative $y$ too, and $\tfrac{d}{dy}\ln(-y) = \tfrac{-1}{-y} = \tfrac1y$ there. Writing $\ln y$ alone would restrict the answer to $y > 0$. Note also that the power rule $\int y^n dy = \tfrac{y^{n+1}}{n+1}$ fails here because $n = -1$ would divide by zero.

4. 4

   Assemble the answer.

   $\int \frac{2\,dy}{-2y} = -\ln|y| + C$

5. 5

   Check by differentiating, and note an equivalent form. $\tfrac{d}{dy}\left(-\ln|y|\right) = -\tfrac1y$, matching the integrand. The result can also be written $\ln\left|\tfrac{1}{y}\right| + C$, since $-\ln|y| = \ln|y|^{-1}$. This integral is exactly the left-hand side that appears when separating the decay equation $\tfrac{dy}{dx} = -2y$.