Evaluate the following integral \[\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}\]

Hint: We start solving this problem by simplifying the numerator using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Then we simplify the expression. Then after simplification, we consider the denominator as some other variable t and change the variables of integral. Then we integrate it with respect to the new variable to get the result. At last, we change that result in terms of the given question to obtain the final answer. We use the formula $\int{\dfrac{1}{x}dx=\ln \left( \left| x \right| \right)+C}$ for integration and the formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$, $\dfrac{d}{dx}\left( \text{constant} \right)=0$ for differentiation while solving the problem.

Complete step by step answer:
Let us start by simplifying the given expression \[\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}\].
Now, let us consider the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
By using the above formula, we write ${{x}^{2}}-1$ as $\left( x+1 \right)\left( x-1 \right)$.
So, we can write it as
$\Rightarrow \int{\dfrac{\left( x+1 \right)\left( x-1 \right)}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}$
Now, we consider the formula ${{a}^{2}}=a\times a$.
By using the above formula, we write ${{\left( x-1 \right)}^{2}}$ as $\left( x-1 \right)\left( x-1 \right)$
Then, we get
$\int{\dfrac{\left( x+1 \right)\left( x-1 \right)}{\left( x-1 \right)\left( x-1 \right)\left( x+3 \right)}dx}$
By cancelling $\left( x-1 \right)$ in numerator and denominator, we get

$\begin{align}
  & \Rightarrow \int{\dfrac{\left( x+1 \right)}{\left( x-1 \right)\left( x+3 \right)}dx} \\
 & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+3x-x-3}dx} \\
 & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}....................\left( 1 \right) \\
\end{align}$

Now, let us consider the quadratic expression ${{x}^{2}}+2x-3$ as some variable $t$.
So, ${{x}^{2}}+2x-3=t...................\left( 2 \right)$
We differentiate both sides of above equation with respective to x, we get
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( {{x}^{2}}+2x-3 \right)=\dfrac{dt}{dx} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( 2x \right)-\dfrac{d}{dx}\left( 3 \right)=\dfrac{dt}{dx} \\
 & \Rightarrow 2x+2-0=\dfrac{dt}{dx} \\
 & \Rightarrow 2x+2=\dfrac{dt}{dx} \\
\end{align}$
Now, we take 2 as common on left hand side of above equation, we get
$2\left( x+1 \right)=\dfrac{dt}{dx}$
$\begin{align}
  & \Rightarrow x+1=\dfrac{dt}{dx}\dfrac{1}{2} \\
 & \Rightarrow \left( x+1 \right)dx=\dfrac{1}{2}dt...............\left( 3 \right) \\
\end{align}$
Now, from equations (1), (2) and (3), we get
$\begin{align}
  & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\int{\dfrac{1}{t}\dfrac{dt}{2}} \\
 & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\int{\dfrac{1}{2t}dt} \\
\end{align}$
As $\dfrac{1}{2}$ is constant, we can write it outside the integral, now we get
$\Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\dfrac{1}{2}\int{\dfrac{1}{t}dt}$
Now, let us consider the formula $\int{\dfrac{1}{x}dx=\ln \left( \left| x \right| \right)+C}$ .
By using the above formula, we can write
$\begin{align}
  & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\dfrac{1}{2}\left( \ln \left( \left| t \right| \right)+C \right) \\
 & \Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\dfrac{1}{2}\ln \left( \left| t \right| \right)+C \\
\end{align}$
Where C is some constant.
Now, by replacing $t$ with ${{x}^{2}}+2x-3$, we get
$\Rightarrow \int{\dfrac{x+1}{{{x}^{2}}+2x-3}dx}=\dfrac{1}{2}\ln \left( \left| {{x}^{2}}+2x-3 \right| \right)+C$
Therefore, \[\int{\dfrac{{{x}^{2}}-1}{{{\left( x-1 \right)}^{2}}\left( x+3 \right)}dx}=\dfrac{1}{2}\ln \left( \left| {{x}^{2}}+2x-3 \right| \right)+C\]

Hence, the answer is \[\dfrac{1}{2}\ln \left( \left| {{x}^{2}}+2x-3 \right| \right)+C\].

Note: There is a possibility of making a mistake while changing the variables in the integral, one might forget to find the value of dx in terms of dt and substitute it in the integral in place of dx. Instead of that they might change the whole expression into a function of t and write dx as dt without substituting the actual value.