Integrals of the Form $\int\dfrac{px+q}{ax^2+bx+c}\,dx$ and $\int\dfrac{px+q}{\sqrt{ax^2+bx+c}}\,dx$

When the numerator is linear, write $px+q=A\cdot\dfrac{d}{dx}(ax^2+bx+c)+B$ for constants $A,B$ found by comparing coefficients. This splits $\displaystyle\int\dfrac{px+q}{ax^2+bx+c}\,dx$ into $A\displaystyle\int\dfrac{\frac{d}{dx}(ax^2+bx+c)}{ax^2+bx+c}\,dx+B\displaystyle\int\dfrac{dx}{ax^2+bx+c}$ — the first piece is a Corollary-III log (put $ax^2+bx+c=t$), the second is the 3.2.4 type. Exactly the same split works for $\displaystyle\int\dfrac{px+q}{\sqrt{ax^2+bx+c}}\,dx$, with the first piece becoming a Corollary-IV square root instead of a log.

$\int\dfrac{2x-3}{3x^2+4x+5}\,dx$. Write $2x-3=A(6x+4)+B$; comparing coefficients, $6A=2\Rightarrow A=\frac13$, and $4A+B=-3\Rightarrow B=-\frac{13}3$. So $I=\frac13\int\dfrac{6x+4}{3x^2+4x+5}dx-\frac{13}3\int\dfrac{dx}{3x^2+4x+5}=\frac13\log(3x^2+4x+5)-\dfrac{13}{3\sqrt{11}}\tan^{-1}\dfrac{3x+2}{\sqrt{11}}+c$ (the second integral evaluated by completing the square, 3.2.4-style).

$\int\sqrt{\dfrac{x-5}{x-7}}\,dx$. Multiply inside the root by $\frac{x-5}{x-5}$: the integrand becomes $\dfrac{x-5}{\sqrt{(x-5)(x-7)}}=\dfrac{x-5}{\sqrt{x^2-12x+35}}$. Write $x-5=\frac12(2x-12)+1$: this splits into $\frac12\int\dfrac{2x-12}{\sqrt{x^2-12x+35}}\,dx+\int\dfrac{dx}{\sqrt{x^2-12x+35}}$. The first is a Corollary-IV square root, $\sqrt{x^2-12x+35}$; the second, after completing the square to $(x-6)^2-1$, is a Formula-5 log. Result: $I=\sqrt{x^2-12x+35}+\log\left|(x-6)+\sqrt{x^2-12x+35}\right|+c$. …