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Rewrite the fraction $\frac{y}{y^2-2y-3}$ inside the integral as the product of two functions: $y\frac{1}{y^2-2y-3}$
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$\int_{4}^{8} y\frac{1}{y^2-2y-3}dy$
Learn how to solve differential calculus problems step by step online. Integrate the function y/(y^2-2y+-3) from 4 to 8. Rewrite the fraction \frac{y}{y^2-2y-3} inside the integral as the product of two functions: y\frac{1}{y^2-2y-3}. We can solve the integral \int y\frac{1}{y^2-2y-3}dy by applying integration by parts method to calculate the integral of the product of two functions, using the following formula. First, identify u and calculate du. Now, identify dv and calculate v.