Integrals — Class 12 Maths Solution

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

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

is an improper fraction, therefore, we convert it into proper fraction by long division method.

$\therefore$ $\cfrac{{{x^2} + x + 1}}{{{x^2} - 1}} = Q + \cfrac{R}{D}$
$\therefore$ $\cfrac{{{x^2} + x + 1}}{{{x^2} - 1}} = x + \cfrac{{2x + 1}}{{{x^2} - 1}}$

….(i)
Now, we write $\cfrac{{2x + 1}}{{{x^2} - 1}} = \cfrac{{2x + 1}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x - 1}}$

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

…..(ii)
Putting $x = - 1$ in (ii),

we get
$- 2 + 1 = A\left( { - 1 - 1} \right)$ $\Rightarrow$ $A = \cfrac{{ - 1}}{{ - 2}} = \cfrac{1}{2}$

Putting $x = - 1$ in (ii),

we get
$2 + 1 = B\left( {1 + 1} \right)$ $\Rightarrow$ $B = \cfrac{3}{2}$
$\therefore$ $\cfrac{{2x + 1}}{{{x^2} - 1}} = \cfrac{1}{{2\left( {x + 1} \right)}} + \cfrac{3}{{2\left( {x - 1} \right)}}$

….(iii)
From (i) and (iii), $\cfrac{{{x^3} + x + 1}}{{{x^2} - 1}} = x + \cfrac{1}{{2\left( {x + 1} \right)}} + \cfrac{3}{{2\left( {x - 1} \right)}}$

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

$= \cfrac{{{x^2}}}{2} + \cfrac{1}{2}\log \left| {x + 1} \right| + \cfrac{3}{2}\log \left| {x - 1} \right| + C$