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

Find the integration of ∫1−x2x(1−2x)dx\int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx} .

Explanation

Solution

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.

Complete step-by-step answer:

Let I=∫1−x2x(1−2x)dxI = \int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx}
Multiply and divide by 2 in the integral we get,
I=∫2−2x22x(1−2x)dx=∫2x2−22x(2x−1)dxI = \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
⇒I=∫(x−22x(2x−1)+12)dx\Rightarrow I = \int {\left( {\dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} + \dfrac{1}{2}} \right)dx}, [∵2x2−22x(2x−1)=x−22x(2x−1)+12]\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,
x−22x(2x−1)=A2x+B2x−1\dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{A}{{2x}} + \dfrac{B}{{2x - 1}}
⇒x−2=A(2x−1)+B(2x)\Rightarrow x - 2 = A\left( {2x - 1} \right) + B\left( {2x} \right)
Now comparing the terms we have,
⇒2A+2B=1\Rightarrow 2A + 2B = 1……………………… (1) (Comparing x terms)
And −A=−2⇒A=2 - A = - 2 \Rightarrow A = 2
So from equation (1) we have,
⇒2(2)+2B=1 ⇒2B=1−4=−3 ⇒B=−32 \begin{gathered} \Rightarrow 2\left( 2 \right) + 2B = 1 \\\ \Rightarrow 2B = 1 - 4 = - 3 \\\ \Rightarrow B = \dfrac{{ - 3}}{2} \\\ \end{gathered}
⇒x−22x(2x−1)=22x+−322x−1\Rightarrow \dfrac{{x - 2}}{{2x\left( {2x - 1} \right)}} = \dfrac{2}{{2x}} + \dfrac{{\dfrac{{ - 3}}{2}}}{{2x - 1}}
So the integral becomes
⇒I=∫(22x+−322x−1+12)dx\Rightarrow I = \int {\left( {\dfrac{2}{{2x}} + \dfrac{{\dfrac{{ - 3}}{2}}}{{2x - 1}} + \dfrac{1}{2}} \right)dx}
⇒I=∫(1x−32(2x−1)+12)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 1x\dfrac{1}{x} is ln x and integration of 1ax+b\dfrac{1}{{ax + b}} is ln⁡(ax+b)a\dfrac{{\ln \left( {ax + b} \right)}}{a} so use these properties we have,
⇒I=ln⁡x−32(ln⁡(2x−1)2)+12(x)+c\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 1x\dfrac{1}{x} is ln x. This always helps after simplification of the integral to get the right answer.