Integrals — Class 12 Maths Solution

.: Let $I = \int {\cfrac{1}{{{x^4} - 1}}dx}$

Let $\cfrac{1}{{{x^4} - 1}} = \cfrac{1}{{\left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^2} + 1} \right)}} = \cfrac{A}{{x + 1}} + \cfrac{B}{{x - 1}} + \cfrac{{Cx + D}}{{{x^2} + 1}}$

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

…(i)
Putting $x = - 1$ in (i),

we get
$1 = A\left( { - 1 - 1} \right)\left( {1 + 1} \right)$ $\Rightarrow$ $1 = A\left( { - 4} \right)$ $\Rightarrow$ $A = - \cfrac{1}{4}$

Putting $x = 1$ in (i),

we get
$1 = B\left( {1 + 1} \right)\left( {1 + 1} \right)$ $\Rightarrow$ $1 = B\left( 2 \right)\left( 2 \right)$ $\Rightarrow$ $B = \cfrac{1}{4}$

Comparing coefficients of ${x^3}$ and constant in (i) on both sides ,

we get
$0 = A + B + C$ and $- A + B - D = 1$

$\Rightarrow$ $0 = \cfrac{{ - 1}}{4} + \cfrac{1}{4} + C$ $\Rightarrow$ $C = 0$ and $1 = \cfrac{1}{4} + \cfrac{1}{4} - D$ $\Rightarrow$ $D = - \cfrac{1}{2}$

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

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

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