Find $\int \frac{y}{y^2 - 4} dy$ $\frac{y^2}{2} + C$ $\frac{\frac{y^3}{3} - 4y}{3} + C$ $ln|2y| + C$ $2ln|y^2 - 4| + C$ $\frac{1}{2}ln|y^2 - 4| + C$ Clear my selection
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Then $du = 2y \, dy$, so $y \, dy = \frac{1}{2} \, du$. Then Show more…
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