Question

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

Answer 1

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 = \int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx} $
Rewrite the above expression by using the concept of algebra of integration as:
\[\int {\dfrac{{1 - {x^2}}}{{x\left( {1 - 2x} \right)}}dx = } \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx - \int {\dfrac{{{x^2}}}{{x\left( {1 - 2x} \right)dx}}} } \]
Cancel out the common factors from numerator and denominator.
$ \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 = \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx} $ and $N = \int {\dfrac{x}{{1 - 2x}}dx} $.
Now simplify $M = \int {\dfrac{1}{{x\left( {1 - 2x} \right)}}dx} $ by using the concept of integration of rational in $M$ we get
$
  1 = A\left( {1 - 2x} \right) + Bx \\
  1 = x\left( { - 2A + B} \right) + A \\
$
On comparing both sides,
$A = 1$ and $ - 2A + B = 0$
Now simplify, $B = 2$ and $A = 2$
Hence,
$M = \int {\dfrac{1}{x} + \dfrac{2}{{1 - 2x}}dx = } \log x = + 2\log \left( {1 - 2x} \right)$.
Now we have to solve $N = \int {\dfrac{x}{{1 - 2x}}dx} $.
Let $1 - 2x = \upsilon $ which implies $x = \dfrac{{\left( {1 - \upsilon } \right)}}{2}$
On differentiating above on both sides, we get $ - 2dx = d\upsilon $
Now substitute $ - 2dx = d\upsilon $ in $N = \int {\dfrac{x}{{1 - 2x}}dx} $ and simplify by integrating,
$N = \dfrac{{ - 1}}{4}\int {1 - \upsilon .d\upsilon = \dfrac{{ - 1}}{4}\left[ {\upsilon - \dfrac{{{\upsilon ^2}}}{2}} \right]} $
Back substitute the value of $\upsilon $ in the above obtained expression.
$
  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, $P = M + N = \log \dfrac{x}{{{{\left( {1 - 2x} \right)}^2}}} + \dfrac{{4{x^2} - 1}}{8} + C$

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.