Question

Evaluate $\int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}}dx$.

Answer 1

Hint: In order to solve this question, we need to separate the variables to integrate and we can do this by \[\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{3}} \right)}=\dfrac{A}{\left( 1-x \right)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}\] .So, our main aim is to find the value of A and B. Further in order to solve the integration w need to know the standard formulas, $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}}dx=\ln \left( f\left( x \right) \right)+C$ and $\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x+C}$.

Complete step by step answer:
We need to evaluate this integral given below,
$\int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}dx}$
In order to solve this, we need to separate the denominators with an additional sign so that we can integrate separately.
We need to be in the form of $\dfrac{A}{\left( 1-x \right)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}$
Therefore, we can write the expression as
\[\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{3}} \right)}=\dfrac{A}{\left( 1-x \right)}+\dfrac{Bx+C}{\left( 1+{{x}^{2}} \right)}\]
Now, our aim is to find the values of A and B.
Let's solve the right-hand side and compare the denominators.
By solving we get,
\[\begin{align}
  & \dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{3}} \right)}=\dfrac{A\left( 1+{{x}^{2}} \right)+\left( Bx+C \right)\left( 1-x \right)}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)} \\
 & \dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{3}} \right)}=\dfrac{A+A{{x}^{2}}+Bx-B{{x}^{2}}+C-Cx}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)} \\
 & \dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{3}} \right)}=\dfrac{\left( A+C \right)+\left( A-B \right){{x}^{2}}+\left( B-C \right)x}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)} \\
\end{align}\]
Now comparing the constant terms of 2 and rest other terms are 0.
Comparing we get,
$\begin{align}
  & A+C=2 \\
 & A-B=0 \\
 & B-C=0 \\
\end{align}$
Solving for A, B, C we get,
$A=B,B=C$
A= 1, B = 1, C = 1.
Hence the integrand becomes as,
$\dfrac{1}{\left( 1-x \right)}+\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}$
Now we can integrate with respect to x.
By integrating we get,
$\begin{align}
  & \int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{1}{\left( 1-x \right)}+\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}dx} \\
 & \int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}dx}=\int{\dfrac{1}{\left( 1-x \right)}dx+\int{\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}}dx} \\
\end{align}$
We need to know the standard differentiation formula to solve this integration.
The standard formulas are $\int{\dfrac{f'\left( x \right)}{f\left( x \right)}}dx=\ln \left( f\left( x \right) \right)+C$ and $\int{\dfrac{1}{1+{{x}^{2}}}={{\tan }^{-1}}x+C}$
Substituting we get,
$\begin{align}
  & \int{\dfrac{1}{\left( 1-x \right)}dx+\int{\dfrac{x+1}{\left( 1+{{x}^{2}} \right)}}dx}=\int{\dfrac{1}{\left( 1-x \right)}dx+\int{\dfrac{x}{1+{{x}^{2}}}dx+\int{\dfrac{1}{1+{{x}^{2}}}dx}}} \\
 & =-\ln \left( 1-x \right)+\ln \left( 1+{{x}^{2}} \right)+{{\tan }^{-1}}x+C
\end{align}$
Hence the integration of the equation $\int{\dfrac{2}{\left( 1-x \right)\left( 1+{{x}^{2}} \right)}}dx$ is $-\ln \left( 1-x \right)+\ln \left( 1+{{x}^{2}} \right)+{{\tan }^{-1}}x+C$ .

Note:
 In this problem, we can see that while integrating we are keeping just a single constant and not all constant from every integration. This is because all the constant from all the integrations add up to a single integration. Also, in the first integration, we introduced a negative sign in the beginning because we already had one negative sign in the denominator.