$\cfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}$

: Let $I = \int {\cfrac{{3x + 5}}{{{x^3} - {x^2} - x + 1}}dx}$

Also, $\cfrac{{3x + 5}}{{{x^2}(x - 1) - 1(x - 1)}} = \cfrac{{3x + 5}}{{\left( {{x^2} - 1} \right)\left( {x - 1} \right)}} = \cfrac{{3x + 5}}{{\left( {x + 1} \right){{\left( {x - 1} \right)}^2}}}$

We write, $\cfrac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}} = \cfrac{A}{{x - 1}} + \cfrac{B}{{{{\left( {x - 1} \right)}^2}}} + \cfrac{C}{{x + 1}}$

$\Rightarrow$ $3x + 5 = A\left( {x - 1} \right)\left( {x + 1} \right) + B\left( {x + 1} \right) + C{\left( {x - 1} \right)^2}$

….(i)
Putting $x = 0$ in $\left( i \right)$ ,

we get
$5 = - A + B + C$

….(ii)
Putting $x = 1$ in $\left( i \right)$ ,

we get
$3 + 5 = B \times 2$ $\Rightarrow$ $B = \cfrac{8}{2} = 4$

Putting $x = - 1$ in $\left( i \right)$ ,

we get
$- 3 + 5 = C{\left( { - 1 - 1} \right)^2}$ $\Rightarrow$ $C = \cfrac{2}{4} = \cfrac{1}{2}$

Now, Putting values of B and C in (ii),

we get
$5 = - A + 4 + \cfrac{1}{2}$ $\Rightarrow$ $A = \cfrac{9}{2} - 5 = - \cfrac{1}{2}$

$\therefore$ $I = \int {\cfrac{{3x + 5}}{{{{\left( {x - 1} \right)}^2}\left( {x + 1} \right)}}dx = - \cfrac{1}{2}\int {\cfrac{{dx}}{{\left( {x - 1} \right)}} + 4} } \int {\cfrac{{dx}}{{{{\left( {x - 1} \right)}^2}}} + \cfrac{1}{2}\int {\cfrac{{dx}}{{x + 1}}} }$

$= \cfrac{1}{2}\log \left| {x - 1} \right| + 4\left( { - \cfrac{1}{{x - 1}}} \right) + \cfrac{1}{2}\log \left| {x + 1} \right| + C$

$= \cfrac{1}{2}\log \left| {\cfrac{{x + 1}}{{x - 1}}} \right| - \cfrac{4}{{x - 1}} + C$

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