How do you integrate $\int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}}dx$ using partial fractions?

Hint: In this question, before using partial fraction, we will use constant factor rule, that is:$\int{cf\left( x \right)dx=c\int{f\left( x \right)}}dx$ and then apply partial fraction and then we will rewrite the integral in partial fractions and then integrate it.

Complete step by step answer:
As we can see in our question, we have to integrate our equation using partial fractions, so first we will use the constant factor rule, which helps us to separate the constant and the equation which we are going to evaluate. Therefore, the constant factor rule is:
$\int{cf\left( x \right)dx=c\int{f\left( x \right)}}dx$
Now, from above question we get:
$2\int{\dfrac{1}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}}dx$
Now here we will begin partial fraction, partial fraction decomposition method is used to decompose a rational expression into simpler fractions; that is if we were given a single complicated fraction, we first have to break it down into a series of “smaller” parts. So we get our equation as:
$\dfrac{1}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}=\dfrac{A}{1-x}+\dfrac{Bx+C}{1+{{x}^{2}}}\to \left( 1 \right)$
Now, Rewrite the right side of the equation with a common denominator, we get:
$\dfrac{1}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}=\dfrac{A\left( 1+{{x}^{2}} \right)+\left( Bx+C \right)\left( 1-x \right)}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}$
Now cancel $\left( 1-x \right)\left( 1+{{x}^{2}} \right)$ we get:
$1=A\left( 1+{{x}^{2}} \right)+\left( Bx+C \right)\left( 1-x \right)$
Now by expanding, we get:
$1=A+A{{x}^{2}}+Bx-B{{x}^{2}}+C-Cx$
Now we have to group terms by degrees of x, we get:
$1=\left( A-B \right){{x}^{2}}+\left( B-C \right)x+A+C$
Now we will set up an equation system based on the degrees of x, we get:
$\begin{align}
  & A-B=0\to \left( 2 \right) \\
 & B-C=0\to \left( 3 \right) \\
 & A+C=1\to \left( 4 \right) \\
\end{align}$
Now in the equation, $\left( 2 \right)$, on transferring $-B$from the left-hand side to the right-hand side, we get:
$A=B$
Now in the equation, $\left( 3 \right)$, on transferring $-C$from the left-hand side to the right-hand side, we get:
$B=C$
Now in equation $\left( 4 \right)$, we have: $A+C=1$
Now since $B=C$, we can substitute it as:
$A+B=1$
Also since $A=B$and therefore on substituting we get:
$A+A=1$
On simplifying, we get:
$2A=1$
$A=\dfrac{1}{2}$
As we can see that, $A=B$ and $B=C$ we get the following values:
$\begin{align}
  & A=\dfrac{1}{2} \\
 & B=\dfrac{1}{2} \\
 & C=\dfrac{1}{2} \\
\end{align}$
Now we will substitute $A,B,C$into the original expression $\left( 1 \right)$, we get:
$\dfrac{1}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}=\dfrac{1}{2\left( 1-x \right)}+\dfrac{x+1}{2\left( 1+{{x}^{2}} \right)}$
With this partial fraction Decomposition completes:
$\dfrac{1}{2\left( 1-x \right)}+\dfrac{x+1}{2\left( 1+{{x}^{2}} \right)}$
Next we will rewrite the integral in partial fractions:
$2\int{\dfrac{1}{2\left( 1-x \right)}+\dfrac{x+1}{2\left( 1+{{x}^{2}} \right)}}$
Now, we will use Sum Rule:
$\int{f\left( x \right)}+g\left( x \right)dx=\int{f\left( x \right)}dx+\int{g\left( x \right)}dx$
Therefore, we get:
$2\left( \int{\dfrac{1}{2\left( 1-x \right)}dx+\int{\dfrac{x+1}{2\left( 1+{{x}^{2}} \right)}dx}} \right)$
Next, use constant factor Rule:
$\int{cf\left( x \right)}dx=c\int{f\left( x \right)}dx$
So we will get:
$2\left( \dfrac{1}{2}\int{\dfrac{1}{\left( 1-x \right)}dx+\int{\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}dx}} \right)$
Now we will use integration by substitution on $\int{\dfrac{1}{1-x}dx}$ we get:
Let $u=1-x,du=-dx$
Next we will substitute $u$and $du$ we get:
$\int{-\dfrac{1}{u}du}$
By taking derivative, we get:
$-\ln u$
Now, substituting $u=1-x$ we get:
$-\ln \left( 1-x \right)$
Now rewrite the integral with the completed substitution:
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\int{\dfrac{x+1}{2\left( 1+{{x}^{2}} \right)}dx} \right)\]
Now use constant factor rule:
$\int{cf\left( x \right)}dx=c\int{f\left( x \right)}dx$
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{1}{2}\int{\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}dx} \right)\]
Now we will split fractions, we get:
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{1}{2}\int{\dfrac{x}{\left( 1+{{x}^{2}} \right)}+\dfrac{1}{1+{{x}^{2}}}dx} \right)\]
Again use Sum Rule:
$\int{f\left( x \right)}+g\left( x \right)dx=\int{f\left( x \right)}dx+\int{g\left( x \right)}dx$
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{1}{2}\left( \int{\dfrac{x}{\left( 1+{{x}^{2}} \right)}dx+\int{\dfrac{1}{1+{{x}^{2}}}dx}} \right) \right)\]
Next we will use integration by substitution on $\int{\dfrac{x}{1+{{x}^{2}}}}dx$
Let $u=1+{{x}^{2}},du=2xdx,$then $xdx=\dfrac{1}{2}du$
Using $u,du$we will rewrite above fraction:
$\int{\dfrac{1}{2u}du}$
$\dfrac{1}{2}\int{\dfrac{1}{u}du}$
Now derivative:
$\dfrac{\ln u}{2}$
Now substituting $u=1+{{x}^{2}}$
$\dfrac{\ln \left( 1+{{x}^{2}} \right)}{2}$
Now rewrite integral with completed substitution, we get:
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{1}{2}\left( \dfrac{\ln \left( 1+{{x}^{2}} \right)}{2}+\int{\dfrac{1}{1+{{x}^{2}}}}dx \right) \right)\]
Let $x=\tan u,dx={{\sec }^{2}}udu$
We get:
$\int{\dfrac{1}{1+{{\left( \tan u \right)}^{2}}}{{\sec }^{2}}udu}$
On simplifying:
$\int{1du}$
Now we will use this rule: $\int{adx=ax+C}$
$u$
We know that:
$u={{\tan }^{-1}}\left( x \right)$
$\Rightarrow {{\tan }^{-1}}\left( x \right)$
Now rewrite integral with completed substitution:
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{\ln \left( 1+{{x}^{2}} \right)}{4}+\dfrac{{{\tan }^{-1}}\left( x \right)}{2} \right)\]
\[2\left( -\dfrac{\ln \left( 1-x \right)}{2}+\dfrac{\ln \left( 1+{{x}^{2}} \right)}{4}+\dfrac{{{\tan }^{-1}}\left( x \right)}{2} \right)+C\]
Hence, the above expression is the final answer.

Note:
It is to be remembered that while doing integration by parts the terms $u$and $v$should follow the sequence of the acronym $ILATE$, which stands for inverse, logarithm, algebraic, trigonometric, and exponential respectively.
It is to be remembered that integration and differentiation are inverses of each other. If the integration of $a$is $b$then the derivative of $b$is $a$.