Let $I=\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}dx$ $=\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx$ Put $6x+7=A\left(\frac{d}{dx}(x^2-9x+20)\right)+B$ $\Rightarrow 6x+7=A(2x-9)+B$ ... (i) Comparing coefficient of x in (i), we get $2A=6\Rightarrow A=3$ Comparing constant terms in (i), we get $7=-9A+B$ $\Rightarrow 7=-9\times 3+B$ $\Rightarrow B=34$ $\therefore I=\int \frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}dx$ Let $I=3I_1+34I_2$ ... (ii) where, $I_1=\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx$Put $x^2-9x+20=t$ $\Rightarrow (2x-9)dx=dt$ $\therefore I_1=\frac{dt}{\sqrt{t}}$ $=2t^{1/2}+C_1$ $=2\sqrt{x^2-9x+20}+C_1$ ... (iii) and, $I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}$ $=\int \frac{dx}{\sqrt{x^2-9x+\left(\frac{9}{2}\right)^2-\left(\frac{9}{2}\right)^2+20}}$ $=\int \frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^2-\frac{81}{4}+20}}$$=\int \frac{dx}{\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}}$$=\log\left|\left(x-\frac{9}{2}\right)+\sqrt{\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right|+C_2$$=\log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^2-9x+20}\right|+C_2$... (iv)From (ii), (iii) and (iv), we get