Integrals — Class 12 Maths Solution

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

We write, $\cfrac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}} = \cfrac{A}{{1 - x}} + \cfrac{{Bx + C}}{{1 + {x^2}}}$

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

…(i)
Putting $x = 1$ in (i),

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

Comparing coefficients of ${x^2}$ on both sides of (i),

we get
$0 = A - B$ $\Rightarrow$ $A = B$ $\Rightarrow$ $B = 1$

…(ii)
Comparing coefficients of constant terms on both sides of (i),

we get
$2 = A + C$ $\Rightarrow$ $C = 2 - 1 = 1$

…(iii)
$\therefore$ $\cfrac{2}{{\left( {1 - x} \right)\left( {1 + {x^2}} \right)}} = \cfrac{1}{{1 - x}} + \cfrac{{x + 1}}{{1 + {x^2}}}$

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

$= \int {\cfrac{1}{{1 - x}}dx} + \int {\cfrac{{x\,dx}}{{1 + {x^2}}}} + \int {\cfrac{1}{{1 + {x^2}}}dx}$

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