Evaluate the expression$\int\limits_{2}^{3}{\dfrac{dx}{1-{{x}^{2}}}}$

a) $\dfrac{1}{2}\log \left( \dfrac{2}{3} \right)$
 b)$2\log \left( \dfrac{2}{3} \right)$
 c)$\dfrac{\pi }{3}$
 d)$\dfrac{\pi }{2}$

Hint: Apply partial fraction decomposition to the given integral and simplify. Then apply \[\int{\dfrac{1}{u}=\log u}\] and integrate, then apply $\log \dfrac{a}{b}=\log a-\log b$ to simplify the answer.

Complete step-by-step answer:
The given expression is $\int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}.dx$.

This can be written as,

$\int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\int\limits_{2}^{3}{\dfrac{1}{(1-x)(1+x)}}dx......(i)

$

Now we will apply partial fraction decomposition, so

$\dfrac{1}{(1-x)(1+x)}=\dfrac{A}{(1-x)}+\dfrac{B}{(1+x)}.........(ii)$

This can be written as,

$1=A(1+x)+B(1-x)$

Substituting \[(x=1)\], we get

$\begin{align}

  & 1=A(1+1)+B(1-1) \\

 & \Rightarrow 1=2A \\

 & \Rightarrow A=\dfrac{1}{2} \\

\end{align}$

Substituting \[(x=-1)\], we get

$\begin{align}

  & 1=A(1-1)+B(1+1) \\

 & \Rightarrow 1=2B \\

 & \Rightarrow B=\dfrac{1}{2} \\

\end{align}$

Substituting the values of ‘A’ and ‘B’ in equation (ii), we get

$\begin{align}

  & \dfrac{1}{(1-x)(1+x)}=\dfrac{\dfrac{1}{2}}{(1-x)}+\dfrac{\dfrac{1}{2}}{(1+x)} \\

 & \Rightarrow \dfrac{1}{(1-x)(1+x)}=\dfrac{1}{2}\left[ \dfrac{1}{(1-x)}+\dfrac{1}{(1+x)} \right] \\

\end{align}$

Substituting this value in equation (i), we get

$\begin{align}

  & \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\int\limits_{2}^{3}{\dfrac{1}{2}\left[ \dfrac{1}{(1-x)}+\dfrac{1}{(1+x)} \right]}dx \\

 & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \int\limits_{2}^{3}{\dfrac{1}{(1-x)}}dx+\int\limits_{2}^{3}{\dfrac{1}{(1+x)}}dx \right] \\

\end{align}$

Let \[u=1-x\Rightarrow du=-dx,v=1+x\Rightarrow dv=dx\] , so the above equation becomes,
$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \int\limits_{2}^{3}{\dfrac{1}{u}}(-du)+\int\limits_{2}^{3}{\dfrac{1}{v}}dv \right]$

We know, \[\int{\dfrac{1}{u}=\log u}\] , so above equation becomes,

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ -\log u+\log v \right]_{2}^{3}$

Substituting back the values of ‘u’ and ‘v’, we get

$\begin{align}

  & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ -\log (1-x)+\log (1+x) \right]_{2}^{3} \\

 & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \log (1+x)-\log (1-x) \right]_{2}^{3} \\

\end{align}$

We know, $\log \dfrac{a}{b}=\log a-\log b$ , so above equation becomes,

$\Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\log \left[ \dfrac{(1+x)}{(1-x)} \right]_{2}^{3}$

Applying the upper and lower bounds, we get

$\begin{align}

  & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \log \dfrac{(1+3)}{(1-3)}-\log \dfrac{(1+2)}{(1-2)} \right] \\

 & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \log \dfrac{4}{2}-\log \dfrac{3}{1} \right] \\

 & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \log 2-\log 3 \right] \\

 & \Rightarrow \int\limits_{2}^{3}{\dfrac{1}{1-{{x}^{2}}}}dx=\dfrac{1}{2}\left[ \log \dfrac{2}{3} \right] \\

\end{align}$

Hence the correct answer is option(a).


Note: The possible mistake is instead of solving by partial fraction decomposition, students consider the formula of $\int{\dfrac{1}{1-{{x}^{2}}}.dx}$.

This will lead to totally different and wrong answers.