How to solve $\int\frac{4y+3}{4y^2-9}dy$?

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For an assignment from my book I have to evaluate: $$ \int\frac{4y+3}{4y^2-9}dy $$ (and therefore actually solve it), but I don't know how to start. Thanks in advance for your help!

For your information, I know the basic principles of the methods partial fractions, short / long division and completing the square.

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Using Partial Fraction Decomposition

$$\dfrac{4y+3}{(2y)^2-3^2}=\dfrac a{2y-3}+\dfrac b{2y+3}$$ where $a,b$ are arbitrary constants

$$\implies4y+3=a(2y+3)+b(2y-3)=2(a+b)y+3(a-b)$$

Compare the constants & the coefficients of $y$ to find $a,b$