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Question: Integrate the rational function\(\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}\)....

Integrate the rational function1−x2x(1−2x)\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}.

Explanation

Solution

Hint: Try to separate different components of the numerator over the common denominator and then integrate it by using the method of partial fraction.

Complete step-by-step solution:
Let P=∫1−x2x(1−2x)dxP = \int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx}
Rewrite the above expression by using the concept of algebra of integration as:
∫1−x2x(1−2x)dx=∫1x(1−2x)dx−∫x2x(1−2x)dx\int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx = } \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx - \int {\dfrac{{{x^2}}}{{x\left( {1 - 2x} \right)dx}}} }
Cancel out the common factors from numerator and denominator.
⇒∫1x(1−2x)dx−∫x1−2xdx\Rightarrow \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx - \int {\dfrac{x}{{1 - 2x}}dx} }
To solve the question in an easy way for that we will name both the integrands separately.
Let,M=∫1x(1−2x)dxM = \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx} and N=∫x1−2xdxN = \int {\dfrac{x}{{1 - 2x}}dx} .
Now simplify M=∫1x(1−2x)dxM = \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx} by using the concept of integration of rational in MM we get
1=A(1−2x)+Bx 1=x(−2A+B)+A  1 = A\left( {1 - 2x} \right) + Bx \\\ 1 = x\left( { - 2A + B} \right) + A \\\
On comparing both sides,
A=1A = 1 and −2A+B=0 - 2A + B = 0
Now simplify, B=2B = 2 and A=2A = 2
Hence,
M=∫1x+21−2xdx=log⁡x=+2log⁡(1−2x)M = \int {\dfrac{1}{x} + \dfrac{2}{{1 - 2x}}dx = } \log x = + 2\log \left( {1 - 2x} \right).
Now we have to solve N=∫x1−2xdxN = \int {\dfrac{x}{{1 - 2x}}dx} .
Let 1−2x=υ1 - 2x = \upsilon which implies x=(1−υ)2x = \dfrac{{\left( {1 - \upsilon } \right)}}{2}
On differentiating above on both sides, we get −2dx=dυ- 2dx = d\upsilon
Now substitute −2dx=dυ- 2dx = d\upsilon in N=∫x1−2xdxN = \int {\dfrac{x}{{1 - 2x}}dx} and simplify by integrating,
N=−14∫1−υ.dυ=−14[υ−υ22]N = \dfrac{{ - 1}}{4}\int {1 - \upsilon .d\upsilon = \dfrac{{ - 1}}{4}\left[ {\upsilon - \dfrac{{{\upsilon ^2}}}{2}} \right]}
Back substitute the value of υ\upsilon in the above obtained expression.
N=−14[1−2x−(1−2x)22] =−14[2−4x−(12−2∗2x∗1+2x2)2] =4x2+1−4x+4x−28 =4x2−18  N = \dfrac{{ - 1}}{4}\left[ {1 - 2x - \dfrac{{{{\left( {1 - 2x} \right)}^2}}}{2}} \right] \\\ = \dfrac{{ - 1}}{4}\left[ {\dfrac{{2 - 4x - \left( {{1^2} - 2*2x*1 + 2{x^2}} \right)}}{2}} \right] \\\ = \dfrac{{4{x^2} + 1 - 4x + 4x - 2}}{8} \\\ = \dfrac{{4{x^2} - 1}}{8} \\\
Hence, P=M+N=log⁡x(1−2x)2+4x2−18+CP = M + N = \log \dfrac{x}{{{{\left( {1 - 2x} \right)}^2}}} + \dfrac{{4{x^2} - 1}}{8} + C

Note: Solving the question becomes easier if different parts are denoted by names like M and N. This approach makes the question easier to handle and solve.