Integrals — Class 12 Maths Solution

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

Put ${x^2} = t$ $\Rightarrow$ $\cfrac{{\left( {t + 1} \right)\left( {t + 2} \right)}}{{\left( {t + 3} \right)\left( {t + 4} \right)}} = \cfrac{{{t^2} + 3t + 2}}{{{t^2} + 7t + 12}}$

$= 1 + \cfrac{{ - 4t - 10}}{{{t^2} + 7t + 12}} = 1 - \left[ {\cfrac{{\left( {4t + 10} \right)}}{{\left( {t + 3} \right)\left( {t + 4} \right)}}} \right]$

We write , $\cfrac{{4t + 10}}{{\left( {t + 3} \right)\left( {t + 4} \right)}} = \cfrac{A}{{t + 3}} + \cfrac{B}{{t + 4}}$

$\Rightarrow$ $4t + 10 = A\left( {t + 4} \right) + B\left( {t + 3} \right)$

….(i)
Putting $t = - 3$ in $(i)$ ,

we get
$4\left( { - 3} \right) + 10 = A\left( { - 3 + 4} \right)$ $\Rightarrow$ $A = - 2$

Putting $t = - 4$ in $(i)$ ,

we get
$4\left( { - 4} \right) + 10 = B\left( { - 4 + 3} \right)$ $\Rightarrow$ $B = 6$

$\therefore$ $\cfrac{{\left( {{x^2} + 1} \right)\left( {{x^2} + 2} \right)}}{{\left( {{x^2} + 3} \right)\left( {{x^2} + 4} \right)}} = 1 - \left[ {\cfrac{{ - 2}}{{t + 3}} + \cfrac{6}{{t + 4}}} \right] = 1 + \left[ {\cfrac{2}{{{x^2} + 3}} - \cfrac{6}{{{x^2} + 4}}} \right]$

$\therefore$ $I = \int {\left[ {1 + \cfrac{2}{{{x^2} + 3}} - \cfrac{6}{{{x^2} + 4}}} \right]} dx$

$= x + \cfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\cfrac{x}{{\sqrt 3 }}} \right) - \cfrac{6}{2}{\tan ^{ - 1}}\left( {\cfrac{x}{2}} \right) + C$

$= x + \cfrac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\cfrac{x}{{\sqrt 3 }} - 3{\tan ^{ - 1}}\cfrac{x}{2} + C$