$\cfrac{{x + 3}}{{{x^2} - 2x - 5}}$

: Let$I = \int {\cfrac{{x + 3}}{{{x^2} - 2x - 5}}dx}$

Let $x + 3 = A\left( {\cfrac{d}{{dx}}\left( {{x^2} - 2x - 5} \right)} \right) + B = A\left( {2x - 2} \right) + B$

….(i)
Comparing the coefficients of $x$ in (i),

we get
$1 = 2A$ $\Rightarrow$ $A = \cfrac{1}{2}$

Comparing the constant terms in (i),

we get
$3 = B - 2A$ $\Rightarrow$ $3 = B - 1$ $\Rightarrow$ $B = 4$

$\therefore$ $I = \int {\cfrac{{\cfrac{1}{2}\left( {2x - 2} \right) + 4}}{{{x^2} - 2x - 5}}} dx$

$= \cfrac{1}{2}\int {\cfrac{{2x - 2}}{{{x^2} - 2x - 5}}dx + 4\int {\cfrac{{dx}}{{{x^2} - 2x - 5}}} }$

$\Rightarrow$ $I = \cfrac{1}{2}{I_1} + 4{I_2}$

….(ii)
Where ${I_1} = \int {\cfrac{{2x - 2}}{{{x^2} - 2x - 5}}} dx$

Put ${x^2} - 2x - 5 = t$ $\Rightarrow$ $\left( {2x - 2} \right)dx = dt$

$\therefore$ ${I_1} = \int {\cfrac{{dt}}{t}} = \log \left| t \right| = \log \left| {{x^2} - 2x - 5} \right| + {C_1}$

….(iii)
And ${I_2} = \int {\cfrac{{dx}}{{{x^2} - 2x - 5}}} = \int {\cfrac{{dx}}{{{{\left( {x - 1} \right)}^2} - 6}}}$

$= \int {\cfrac{{dx}}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}} = \cfrac{1}{{2\sqrt 6 }}\log \left| {\cfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right| + {C_2}$

….(iv)
Hence, from $(ii),(iii)$ and $(iv)$,

we get
$I = \cfrac{1}{2}\log \left| {\left( {{x^2} - 2x - 5} \right)} \right| + \cfrac{2}{{\sqrt 6 }}\log \left| {\cfrac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right| + C$

$\left[ {C = \cfrac{1}{2}{C_1} + 4{C_2}} \right]$