Integrate the function: $\frac{6x+7}{\sqrt{(x-5)(x-4)}}$

(N/A) We have $\frac{6x+7}{\sqrt{(x-5)(x-4)}} = \frac{6x+7}{\sqrt{x^2-9x+20}}$.
Let $6x+7 = A\frac{d}{dx}(x^2-9x+20) + B$.
$6x+7 = A(2x-9) + B$.
Equating the coefficients of $x$ and the constant term,we get $2A = 6 \Rightarrow A = 3$ and $-9A + B = 7 \Rightarrow -27 + B = 7 \Rightarrow B = 34$.
Thus,$\int \frac{6x+7}{\sqrt{x^2-9x+20}} dx = \int \frac{3(2x-9) + 34}{\sqrt{x^2-9x+20}} dx = 3 \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx + 34 \int \frac{1}{\sqrt{x^2-9x+20}} dx$.
Let $I_1 = \int \frac{2x-9}{\sqrt{x^2-9x+20}} dx$. Substituting $t = x^2-9x+20$,$dt = (2x-9)dx$,we get $I_1 = \int t^{-1/2} dt = 2\sqrt{t} = 2\sqrt{x^2-9x+20}$.
Let $I_2 = \int \frac{1}{\sqrt{x^2-9x+20}} dx$. Completing the square,$x^2-9x+20 = (x-\frac{9}{2})^2 - \frac{1}{4} = (x-\frac{9}{2})^2 - (\frac{1}{2})^2$.
Using the formula $\int \frac{1}{\sqrt{x^2-a^2}} dx = \log|x + \sqrt{x^2-a^2}|$,we get $I_2 = \log|x-\frac{9}{2} + \sqrt{x^2-9x+20}|$.
Combining these,the integral is $6\sqrt{x^2-9x+20} + 34\log|x-\frac{9}{2} + \sqrt{x^2-9x+20}| + C$.