Question

Find the integration of $\int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx}$.

Hint: In this question first of all multiply and divide with 2 both in numerator and denominator to break the integration into a simplified form. Use the concept of partial fractions to resolve the integration further to simply it.

Let $I = \int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx}$
Multiply and divide by 2 in the integral we get,
$I = \int {\dfrac{{2 - 2{x^2}}}{{2x\left( {1 - 2x} \right)}}dx} = \int {\dfrac{{2{x^2} - 2}}{{2x\left( {2x - 1} \right)}}dx}$
Now the above integral is also written as
$\Rightarrow I = \int {\left( {\dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} + \dfrac{1}{2}} \right)dx}$, $\left[ {\because \dfrac{{2{x^2} - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} + \dfrac{1}{2}} \right]$
Apply partial fraction
So let,
$\dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{A}{{2x}} + \dfrac{B}{{2x - 1}}$
$\Rightarrow x - 2 = A\left( {2x - 1} \right) + B\left( {2x} \right)$
Now comparing the terms we have,
$\Rightarrow 2A + 2B = 1$……………………… (1) (Comparing x terms)
And $- A = - 2 \Rightarrow A = 2$
So from equation (1) we have,
$\begin{gathered} \Rightarrow 2\left( 2 \right) + 2B = 1 \\ \Rightarrow 2B = 1 - 4 = - 3 \\ \Rightarrow B = \dfrac{{ - 3}}{2} \\ \end{gathered}$
$\Rightarrow \dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{2}{{2x}} + \dfrac{{\dfrac{{ - 3}}{2}}}{{2x - 1}}$
So the integral becomes
$\Rightarrow I = \int {\left( {\dfrac{2}{{2x}} + \dfrac{{\dfrac{{ - 3}}{2}}}{{2x - 1}} + \dfrac{1}{2}} \right)dx}$
$\Rightarrow I = \int {\left( {\dfrac{1}{x} - \dfrac{3}{{2\left( {2x - 1} \right)}} + \dfrac{1}{2}} \right)dx}$
Now integrate the above equation we have, as we know integration of $\dfrac{1}{x}$ is ln x and integration of $\dfrac{1}{{ax + b}}$ is $\dfrac{{\ln \left( {ax + b} \right)}}{a}$ so use these properties we have,
$\Rightarrow I = \ln x - \dfrac{3}{2}\left( {\dfrac{{\ln \left( {2x - 1} \right)}}{2}} \right) + \dfrac{1}{2}\left( x \right) + c$, where c is some arbitrary integration constant.
So this is the required value of the integration.

Note: Whenever we face such types of problems the key concept is to have a good grasp over the standard integration form just like integration of $\dfrac{1}{x}$ is ln x. This always helps after simplification of the integral to get the right answer.