Integrals — Maths STD 12 Science — Question

$= \frac{1}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}$

Putting ${x^2} = y$,we get,

$\frac{1}{{{x^4} - 1}}$$=\frac{1}{{\left( {y - 1} \right)\left( {y + 1} \right)}} $

$Let\ \frac{1}{{\left( {y - 1} \right)\left( {y + 1} \right)}} = \frac{A}{{y - 1}} + \frac{B}{{y + 1}}$ .....(i)

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

$ \Rightarrow 1 = Ay + A + By - B$

Comparing the coefficients of y, A + B = 0 ......(ii)

Comparing constants A – B = 1 .......(iii)

On solving the eq. (ii) and (iii), we get $A = \frac{1}{2},B = \frac{{ - 1}}{2}$

Putting the values of A, B and y in eq. (i),

$\frac{1}{{{x^4} - 1}} = \frac{{\frac{1}{2}}}{{{x^2} - 1}} + \frac{{\frac{{ - 1}}{2}}}{{{x^2} + 1}}$

$\Rightarrow \int {\frac{1}{{{x^4} - 1}}dx = \frac{1}{2}\int {\frac{1}{{{x^2} - 1}}dx - \frac{1}{2}\int {\frac{1}{{{x^2} + 1}}dx} } } $

$= \frac{1}{2}.\frac{1}{{2.1}}\log \left| {\frac{{x - 1}}{{x + 1}}} \right| - \frac{1}{2}{\tan ^{ - 1}}x + c$

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