Q.Evaluate: $\int \frac{2x+3}{2x^2+3x-1}\,dx$

Derivative of denominator is $4x+3$. Write $2x+3=A(4x+3)+B$: $4A=2\Rightarrow A=\dfrac12$, $3A+B=3\Rightarrow B=\dfrac32$.

So $\displaystyle\int\dfrac{2x+3}{2x^2+3x-1}dx=\dfrac12\int\dfrac{4x+3}{2x^2+3x-1}dx+\dfrac32\int\dfrac{dx}{2x^2+3x-1}$.

First: $\dfrac12\log\left|2x^2+3x-1\right|$.

Second: factor out 2 and complete the square, $2x^2+3x-1=2\left[\left(x+\dfrac34\right)^2-\dfrac{17}{16}\right]$, giving $\dfrac{1}{\sqrt{17}}\log\left|\dfrac{4x+3-\sqrt{17}}{4x+3+\sqrt{17}}\right|$, so the $\dfrac32$ term contributes $\dfrac{3}{2\sqrt{17}}\log\left|\dfrac{4x+3-\sqrt{17}}{4x+3+\sqrt{17}}\right|$.

Split numerator into derivative-of-denominator + constant, then log + completing-the-square log forms

Forgetting to factor the leading coefficient 2 out before completing the square, giving a wrong value of $a$ under the square root.