. $\cfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}$

: Let $I = \int {\cfrac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}} dx = \int {\cfrac{{\left( {6x + 7} \right)dx}}{{\sqrt {{x^2} - 9x + 20} }}}$

Let $6x + 7 = A \times \left[ {\cfrac{d}{{dx}}\left( {{x^2} - 9x + 20} \right)} \right] + B$

$\Rightarrow$ $6x + 7 = A\left( {2x - 9} \right) + B$

….(i)
Comparing coefficients of x in (i), we get $2A = 6$ $\Rightarrow$ $A = 3$

Comparing constant terms in (i),

we get
$7 = - 9A + B$ $\Rightarrow$ $7 = - 9 \times 3 + B$ $\Rightarrow$ $B = 34$

$\therefore$ $I = \int {\cfrac{{3\left( {2x - 9} \right) + 34}}{{\sqrt {{x^2} - 9x + 20} }}dx}$

$I = 3\int {\cfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx + 34\int {\cfrac{{dx}}{{\sqrt {{x^2} - 9x + 20} }}} }$

$I = 3{I_1} + 34{I_2}$

(Say)

…….(ii)
Let ${I_1} = \int {\cfrac{{2x - 9}}{{\sqrt {{x^2} - 9x + 20} }}dx} = \int {\cfrac{{dt}}{t} = \int {{t^{1/2}}dt} = 2{t^{1/2}} + {C_1}}$

$= 2\sqrt {{x^2} - 9x + 20} + {C_1}$

…..(iii)
${I_2} = \int {\cfrac{{dx}}{{\sqrt {{x^2} - 9x + 20} }}} = \int {\cfrac{{dx}}{{\sqrt {{x^2} - 9x + {{\left( {\cfrac{9}{2}} \right)}^2} - {{\left( {\cfrac{9}{2}} \right)}^2} + 20} }}}$

${I_2} = \int {\cfrac{{dx}}{{\sqrt {{{\left( {x - \cfrac{9}{2}} \right)}^2} - \cfrac{{81}}{4} + 20} }}} = \int {\cfrac{{dx}}{{\sqrt {{{\left( {x - \cfrac{9}{2}} \right)}^2} - {{\left( {\cfrac{1}{2}} \right)}^2}} }}}$

$= \log \left| {x - \cfrac{9}{2} + \sqrt {{{\left( {x - \cfrac{9}{2}} \right)}^2} - {{\left( {\cfrac{1}{2}} \right)}^2}} } \right| + {C_2}$

$= \log \left| {\left( {x - \cfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right| + {C_2}$

…..(iv)
From (ii),(iii) and (iv)

we get
$I = 3 \times 2\sqrt {{x^2} - 9x + 20} + 34\log \left| {\left( {x - \cfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right| + C$

Or $I = 6\sqrt {{x^2} - 9x + 20} + 34\log \left| {\left( {x - \cfrac{9}{2}} \right) + \sqrt {{x^2} - 9x + 20} } \right| + C$

$\left[ {{C_1} = 3{C_1} + 34{C_2}} \right]$