Question

Integrate the following expression $\int {\dfrac{{dx}}{{x - {x^3}}}}$.
A) $\dfrac{1}{2}\log \left| {\dfrac{{1 - {x^2}}}{{{x^2}}}} \right| + C$
B) $\log \left| {\dfrac{{1 - x}}{{x(1 + x)}}} \right| + C$
C) $\log \left| {x - {x^3}} \right| + C$
D) $\dfrac{1}{2}\log \left| {\dfrac{{{x^2}}}{{1 - {x^2}}}} \right| + C$

Answer 1

We can solve this integral using the method of partial fraction. All we need is to split into simplest fractions so that we can integrate directly. Then applying some logarithmic formulas, we get the answer.

Formula used: For every $x,y$, we have, ${x^2} - {y^2} = (x - y)(x + y)$
For every function $f(x)$, $\int {\dfrac{1}{{f(x)}}} dx = \log (f(x))\dfrac{{df(x)}}{{dx}} + C$, where $C$ is the constant of integration.
Also, $\log {x^2} = 2\log x,\log (A + B) = \log A\log B,\log (A - B) = \dfrac{{\log A}}{{\log B}}$, for every function $A,B$.

Complete step-by-step answer:
The given function is $\dfrac{1}{{x - {x^3}}}$.
Taking $x$ common from the denominator we have,
$\Rightarrow \dfrac{1}{{x - {x^3}}} = \dfrac{1}{{x(1 - {x^2})}}$
For every $x,y$, we have, ${x^2} - {y^2} = (x - y)(x + y)$
Applying this result,
$\Rightarrow \dfrac{1}{{x - {x^3}}} = \dfrac{1}{{x(1 - {x^2})}} = \dfrac{1}{{x(1 - x)(1 + x)}} - - - (i)$
Suppose there exists $A,B,C$ such that
$\Rightarrow \dfrac{1}{{x(1 - x)(1 + x)}} = \dfrac{A}{x} + \dfrac{B}{{1 - x}} + \dfrac{C}{{1 + x}} - - - (ii)$
Taking LCM on Right hand side and rearranging we have,
$\Rightarrow \dfrac{1}{{x(1 - x)(1 + x)}} = \dfrac{{A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x)}}{{x(1 - x)(1 + x)}}$
Comparing the sides of the above equation, we have,
$\Rightarrow 1 = A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x)$
Simplifying we get,
$\Rightarrow 1 = A(1 - {x^2}) + B(x + {x^2}) + C(x - {x^2})$
Splitting into terms of $x,{x^2}$ and constant,
$\Rightarrow 1 + 0x + 0{x^2} = A + ( - A + B - C){x^2} + (B + C)x$
Comparing both sides of the above equation we have,
$\Rightarrow A = 1, - A + B - C = 0,B + C = 0$
Substituting we get,
$- 1 + B - C = 0 \Rightarrow B - C = 1$
$B + C = 0 \Rightarrow B = - C$
Again, substituting we get,
$B + B = 1 \Rightarrow 2B = 1 \Rightarrow B = \dfrac{1}{2}$
Since $B = \dfrac{1}{2},B = - C \Rightarrow C = - \dfrac{1}{2}$
So, we have, $A = 1,B = \dfrac{1}{2},C = - \dfrac{1}{2}$
Substituting for $A,B,C$ in equation $(ii)$,
$\Rightarrow \dfrac{1}{{x(1 - x)(1 + x)}} = \dfrac{A}{x} + \dfrac{B}{{1 - x}} + \dfrac{C}{{1 + x}} = \dfrac{1}{x} + \dfrac{1}{{2(1 - x)}} - \dfrac{1}{{2(1 + x)}} - - (iii)$
We have to find $\int {\dfrac{1}{{x - {x^3}}}} dx$
From $(i),(iii)$ we have,
$\int {\dfrac{{dx}}{{x - {x^3}}}} = \int {[\dfrac{1}{x} + \dfrac{1}{2}\dfrac{1}{{(1 - x)}}} - \dfrac{1}{2}\dfrac{1}{{(1 + x)}}]dx$
$\Rightarrow \int {\dfrac{{dx}}{{x - {x^3}}}} = \int {\dfrac{1}{x}dx + \int {} \dfrac{1}{2}\dfrac{1}{{(1 - x)}}} dx - \int {} \dfrac{1}{2}\dfrac{1}{{(1 + x)}}dx$
For every function $f(x)$, $\int {\dfrac{1}{{f(x)}}} dx = \log (f(x))\dfrac{{df(x)}}{{dx}} + C$, where $C$ is the constant of integration.
Here letting $f(x)$ as $1 - x,1 + x$, we have,
$\int {\dfrac{{dx}}{{x - {x^3}}}} = \log x - \dfrac{1}{2}\log (1 - x) - \dfrac{1}{2}\log (1 + x) + C$, where $C$ is the constant of integration.
$\Rightarrow \int {\dfrac{{dx}}{{x - {x^3}}}} = \log x - \dfrac{1}{2}[\log (1 - x) + \log (1 + x)] + C$
We know $\log {x^2} = 2\log x,\log (A + B) = \log A\log B,\log (A - B) = \dfrac{{\log A}}{{\log B}}$
$\Rightarrow \int {\dfrac{{dx}}{{x - {x^3}}}} = \dfrac{1}{2}\log {x^2} - \dfrac{1}{2}[\log (1 - x)(1 + x)] + C$
Since $(1 - x)(1 + x) = 1 - {x^2}$and simplifying,
$\Rightarrow \int {\dfrac{{dx}}{{x - {x^3}}}} = \dfrac{1}{2}[\log {x^2} - \log (1 - {x^2})] + C$
Applying the result $\log A - \log B = \log \dfrac{A}{B}$, in the above equation,
$\Rightarrow \int {\dfrac{{dx}}{{x - {x^3}}}} = \dfrac{1}{2}\log \dfrac{{{x^2}}}{{1 - {x^2}}} + C$, where $C$ is the constant of integration.

The answer is $\dfrac{1}{2}\log \left| {\dfrac{{{x^2}}}{{1 - {x^2}}}} \right| + C$.

So, the correct answer is “Option D”.

Note: We must be careful while applying the results of integration and logarithm. This question may be solved in other ways instead of partial fraction. Since the integral used here is the indefinite one it is mandatory to add the constant of integration.Students should remember the formulas of integration , differentiation and logarithmic properties for solving these types of problems and also should know the method of solving integration by partial fraction method.