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

Let $Q=-2x^2+x+3$, so $Q'=-4x+1$. Write $3x+1=A\cdot Q'+B=A(-4x+1)+B$: matching coefficients, $-4A=3\Rightarrow A=-\frac34$, and $A+B=1\Rightarrow B=\frac74$. So $\int\frac{3x+1}{\sqrt Q}dx=-\frac34\int\frac{Q'}{\sqrt Q}dx+\frac74\int\frac{dx}{\sqrt Q}$. The first piece is $-\frac34\cdot2\sqrt Q=-\frac32\sqrt Q$. For the second, complete the square: $Q=-2x^2+x+3=2\left[\frac{25}{16}-\left(x-\frac14\right)^2\right]$, so $\int\frac{dx}{\sqrt Q}=\frac{1}{\sqrt2}\sin^{-1}\left(\frac{x-1/4}{5/4}\right)+c=\frac{1}{\sqrt2}\sin^{-1}\left(\frac{4x-1}{5}\right)+c$. So the second piece is $\frac74\cdot\frac{1}{\sqrt2}\sin^{-1} …

Split the linear numerator into (multiple of derivative of the radicand) + constant, s …

Sign error completing the square for a negative leading coefficient, or misplacing the factor of 2 pulled out front which …