Integrals — Class 12 Maths Solution

.: Let $I = \int {\cfrac{1}{{x\left( {{x^4} - 1} \right)}}} dx = \cfrac{1}{4}\int {\cfrac{{4{x^3}dx}}{{{x^4}\left( {{x^4} - 1} \right)}}}$

Put ${x^4} = t$ $\Rightarrow$ $4{x^3}dx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\cfrac{{dt}}{{t\left( {t - 1} \right)}}}$

We write, $\cfrac{1}{{t\left( {t - 1} \right)}} = \cfrac{A}{t} + \cfrac{B}{{t - 1}}$
$\Rightarrow$ $1 = A\left( {t - 1} \right) + Bt$

…(i)
Putting $t = 0$ in $(i)$ ,

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

Putting $t = 1$ in $(i)$ ,

we get $1 = B\left( 1 \right)$ $\Rightarrow$ $B = 1$
$\therefore$ $\cfrac{1}{{t\left( {t - 1} \right)}} = \cfrac{{ - 1}}{t} + \cfrac{1}{{t - 1}}$

$\Rightarrow$ $I = \cfrac{1}{4}\int {\left( {\cfrac{{ - 1}}{t} + \cfrac{1}{{t - 1}}} \right)} dt = \cfrac{1}{4}\left[ { - \log \left| t \right| + \log \left| {t - 1} \right|} \right] + C$

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

********************************
**************************