Integrals — Class 12 Maths Solution

: Since $\cfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}} = \cfrac{{1 - {x^2}}}{{x - 2{x^2}}}$ is an improper fraction, therefore,

we convert it into a proper fraction by long division
method .

we get $\cfrac{{{x^2} - 1}}{{2{x^2} - x}} = \cfrac{1}{2} + \cfrac{{\cfrac{x}{2} - 1}}{{2{x^2} - x}}$

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

Now, $\cfrac{{x - 2}}{{2{x^2} - x}} = \cfrac{{x - 2}}{{x\left( {2x - 1} \right)}} = \cfrac{A}{x} + \cfrac{B}{{2x - 1}}$

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

…(i)
Putting $x = 0$ in $(i)$,

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

Putting $x = \cfrac{1}{2}$ in $(i)$,

we get
$\cfrac{1}{2} - 2 = B\left( {\cfrac{1}{2}} \right)$ $\Rightarrow$ $1 - 4 = B$ $\Rightarrow$ $B = - 3$

$\therefore$ $\cfrac{{x - 2}}{{2{x^2} - x}} = \cfrac{2}{x} - \cfrac{3}{{2x - 1}} = \cfrac{2}{x} + \cfrac{3}{{1 - 2x}}$

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

$= \cfrac{1}{2}x + \log \left| x \right| - \cfrac{3}{4}\log \left| {1 - 2x} \right| + C$