Given:$I=\int \frac{6x+7}{(x-5)(x-4)}dx=\int\frac{6x+7}{\sqrt{x^2-9x+20}}.$
We can write,
6x+7=$A\frac{d}{dx}(x^2-9x+20)+B.$
$\;\;\;=A(2x-9)+B.$
Now equating the coefficients we get,
6=2A$\Rightarrow A=3.$
$7=-9A+B\Rightarrow B=34.$
Hence A=3 and B=34.
$I=\int\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}$.
On separating we can write,
$I=3\int\frac{(2x-9)dx}{\sqrt{x^2-9x+20}}+34\int\frac{dx}{\sqrt{x^2-9x+20}}.$
Let $x^2-9x+20=t.$ and $x^2-9x+20=(x-\frac{9}{2})^2-\frac{1}{4}=(x-\frac{9}{2})^2-(\frac{1}{2})^2$.
On differentiating
$(2x-9)dx=dt.$
Therefore $I=3\int\frac{dt}{\sqrt t}+34\int\frac{dx}{\sqrt{(x-\frac{9}{2}}})^2-(\frac{1}{2})^2.$
On integrating we get,
$\;\;\;=3\begin{bmatrix}\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}\end{bmatrix}+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]$.
$\;\;\;=3\times 2(\sqrt t)+34log|(x-9/2)+\sqrt{(x-5)(x-4)}+c.$
Substituting for t we get,
$\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}dx=6(\sqrt{x^2-9x+2})+34[log|(x-9/2)+\sqrt{(x-5)(x-4)}|]+c.$