How do you integrate $\int{\dfrac{1}{{{x}^{4}}-16}dx}$ using partial fractions?

Hint: To integrate the above integration, we are going to first of all factorize the denominator of the above fraction. The factorization can be done using the following algebraic property i.e. $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$. To integrate the above fraction, we are going to represent his integration in the following form: $\dfrac{1}{ax+b}+\dfrac{1}{cx+d}+\dfrac{ex+f}{g{{x}^{2}}+h}$.

Complete step by step answer:
In the above problem, we are asked to integrate the following using partial fraction:
$\int{\dfrac{1}{{{x}^{4}}-16}dx}$
As you can see that the denominator of the above fraction is of the form $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ so applying this identity in the above we get,
$\begin{align}
  & \int{\dfrac{1}{{{\left( {{x}^{2}} \right)}^{2}}-{{\left( 4 \right)}^{2}}}dx} \\
 & =\int{\dfrac{1}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+4 \right)}dx} \\
 & =\int{\dfrac{1}{\left( {{\left( x \right)}^{2}}-{{\left( 2 \right)}^{2}} \right)\left( {{x}^{2}}+4 \right)}dx} \\
\end{align}$
Now, we can use the above identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a-b \right)\left( a+b \right)$ in ${{\left( {{x}^{2}} \right)}^{2}}-{{\left( 2 \right)}^{2}}$ then we get,
$\int{\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}dx}$
Using partial fractions, the above fraction is equal to:
$\dfrac{A}{x-2}+\dfrac{B}{x+2}+\dfrac{Cx+D}{{{x}^{2}}+4}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}$
Taking L.C.M of the denominator in the L.H.S of the above equation we get,
$\dfrac{A\left( x+2 \right)\left( {{x}^{2}}+4 \right)+B\left( x-2 \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right)\left( x-2 \right)\left( x+2 \right)}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}$
Denominator in the L.H.S and R.H.S will be cancelled out and the above equation will look like:
$\begin{align}
  & A\left( x+2 \right)\left( {{x}^{2}}+4 \right)+B\left( x-2 \right)\left( {{x}^{2}}+4 \right)+\left( Cx+D \right)\left( x-2 \right)\left( x+2 \right)=1 \\
 & \Rightarrow A\left( {{x}^{3}}+8+4x+2{{x}^{2}} \right)+B\left( {{x}^{3}}-8-2{{x}^{2}}+4x \right)+\left( Cx+D \right)\left( {{x}^{2}}-4 \right)=1 \\
 & \Rightarrow A\left( {{x}^{3}}+8+4x+2{{x}^{2}} \right)+B\left( {{x}^{3}}-8-2{{x}^{2}}+4x \right)+\left( C{{x}^{3}}-4Cx+D{{x}^{2}}-4D \right)=1 \\
 & \Rightarrow \left( A+B+C \right){{x}^{3}}+\left( 2A-2B+D \right){{x}^{2}}+\left( 4A+4B-4C \right)x+8A-8B-4D=1+0.{{x}^{3}}+0.{{x}^{2}}+0.x \\
\end{align}$
Comparing the coefficients of ${{x}^{3}},{{x}^{2}},x$ and constant on both the sides we get,
$\begin{align}
  & A+B+C=0......Eq.(1) \\
 & 2A-2B+D=0.......Eq.(2) \\
 & 4A+4B-4C=0........Eq.(3) \\
 & 8A-8B-4D=1........Eq.(4) \\
\end{align}$
In eq. (3), taking 4 as common from L.H.S we get,
$\begin{align}
  & 4\left( A+B-C \right)=0 \\
 & \Rightarrow A+B-C=0.......Eq.(5) \\
\end{align}$
Adding eq. (1) and eq. (5) we get,
$\begin{align}
  & A+B+C=0 \\
 & \dfrac{A+B-C=0}{2A+2B+0=0} \\
\end{align}$
Rewriting the above equation we get,
$\begin{align}
  & 2A+2B=0 \\
 & \Rightarrow 2\left( A+B \right)=0 \\
 & \Rightarrow A+B=0 \\
 & \Rightarrow A=-B.......Eq.(6) \\
\end{align}$
Substituting the above relation in A and B in eq. (2) and eq. (4) we get,
$\begin{align}
  & 2\left( -B \right)-2B+D=0 \\
 & \Rightarrow -4B+D=0 \\
 & \Rightarrow 4B=D.......Eq.(7) \\
\end{align}$
$\begin{align}
  & 8A-8B-4D=1 \\
 & \Rightarrow 8\left( -B \right)-8B-4D=1 \\
 & \Rightarrow -16B-4D=1 \\
 & \Rightarrow -4\left( 4B+D \right)=1 \\
\end{align}$
Using eq. (7) in the above equation we get,
$\begin{align}
  & -4\left( D+D \right)=1 \\
 & \Rightarrow -8D=1 \\
 & \Rightarrow D=-\dfrac{1}{8} \\
\end{align}$
Substituting the above value of D in eq. (7) we get,
$\begin{align}
  & 4B=D \\
 & \Rightarrow 4B=-\dfrac{1}{8} \\
\end{align}$
Dividing 4 on both the sides of the above equation we get,
$B=-\dfrac{1}{32}$
Substituting the above value of B in eq. (6) we get,
$\begin{align}
  & A=-\left( -\dfrac{1}{32} \right) \\
 & \Rightarrow A=\dfrac{1}{32} \\
\end{align}$
Substituting the above value of A and B in eq. (1) we get,
$\begin{align}
  & A+B+C=0 \\
 & \Rightarrow \dfrac{1}{32}-\dfrac{1}{32}+C=0 \\
 & \Rightarrow C=0 \\
\end{align}$
From the above calculations, we have calculated the above values of A, B, C and D. So, substituting these values in $\dfrac{A}{x-2}+\dfrac{B}{x+2}+\dfrac{Cx+D}{{{x}^{2}}+4}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)}$ we get,
$\begin{align}
  & \dfrac{A}{x-2}+\dfrac{B}{x+2}+\dfrac{Cx+D}{{{x}^{2}}+4}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)} \\
 & \Rightarrow \dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}+\dfrac{\left( 0 \right)x-\dfrac{1}{8}}{{{x}^{2}}+4}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)} \\
 & \Rightarrow \dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)}=\dfrac{1}{\left( x-2 \right)\left( x+2 \right)\left( {{x}^{2}}+4 \right)} \\
\end{align}$
Substituting L.H.S of the above equation in place of the above fraction in the given integration we get,
$\int{\left( \dfrac{1}{32\left( x-2 \right)}-\dfrac{1}{32\left( x+2 \right)}-\dfrac{1}{8\left( {{x}^{2}}+4 \right)} \right)}dx$
Rearranging the above equation we get,
$\int{\dfrac{1}{32\left( x-2 \right)}dx}-\int{\dfrac{1}{32\left( x+2 \right)}dx}-\int{\dfrac{1}{8\left( {{x}^{2}}+4 \right)}dx}$
Solving the integration $\int{\dfrac{1}{8\left( {{x}^{2}}+4 \right)}dx}$ by taking $x=2\tan \theta $ in the denominator and we get,
$x=2\tan \theta $
Differentiating on both the sides we get,
$dx=2{{\sec }^{2}}\theta d\theta $
Substituting the value of x and dx in the above integration we get,
$\begin{align}
  & \int{\dfrac{1}{8\left( {{\left( 2\tan \theta \right)}^{2}}+4 \right)}\left( 2{{\sec }^{2}}\theta d\theta \right)} \\
 & =\int{\dfrac{1}{8\left( 4{{\tan }^{2}}\theta +4 \right)}}\left( 2{{\sec }^{2}}\theta d\theta \right) \\
\end{align}$
Now, taking 4 as common from the denominator we get,
$\int{\dfrac{1}{8\left( 4 \right)\left( {{\tan }^{2}}\theta +1 \right)}}\left( 2{{\sec }^{2}}\theta d\theta \right)$
We know that the trigonometric identity is equal to:
$1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $
Using the above relation in $\int{\dfrac{1}{8\left( 4 \right)\left( {{\tan }^{2}}\theta +1 \right)}}\left( 2{{\sec }^{2}}\theta d\theta \right)$ we get,
$\begin{align}
  & \int{\dfrac{1}{8\left( 4 \right)\left( {{\sec }^{2}}\theta \right)}}\left( 2{{\sec }^{2}}\theta d\theta \right) \\
 & =\int{\dfrac{1}{16}}\left( d\theta \right) \\
\end{align}$
Integrating the above we get,
$\dfrac{\theta }{16}+C$
In the above, we have assumed $x=2\tan \theta $ so rearranging this equation so that we will get $\theta $ in terms of x and we get,
$\dfrac{x}{2}=\tan \theta $
Taking ${{\tan }^{-1}}$ on both the sides we get,
${{\tan }^{-1}}\left( \dfrac{x}{2} \right)=\theta $
Substituting the above value of $\theta $ in $\dfrac{\theta }{16}+C$ we get,
$\dfrac{{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{16}+C$
Hence, we have integrated $\int{\dfrac{1}{8\left( {{x}^{2}}+4 \right)}dx}$ to $\dfrac{{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{16}+C$.
Now, we are going to do the integration of $\int{\dfrac{1}{32\left( x-2 \right)}dx}-\int{\dfrac{1}{32\left( x+2 \right)}dx}$. We know the integration of:
$\int{\dfrac{dx}{ax+b}=\dfrac{1}{a}}\ln \left| ax+b \right|+C$
So, using the above integration in integrating $\int{\dfrac{1}{32\left( x-2 \right)}dx}-\int{\dfrac{1}{32\left( x+2 \right)}dx}$ we get,
$\dfrac{1}{32}\ln \left| x-2 \right|-\dfrac{1}{32}\ln \left| x+2 \right|+C$
And in the above, we have integrated $-\int{\dfrac{1}{8\left( {{x}^{2}}+4 \right)}dx}=-\dfrac{1}{16}{{\tan }^{-1}}\left( \dfrac{x}{2} \right)+C$

Hence, we have integrated the given integration to:
$\dfrac{1}{32}\ln \left| x-2 \right|-\dfrac{1}{32}\ln \left| x+2 \right|-\dfrac{{{\tan }^{-1}}\left( \dfrac{x}{2} \right)}{16}+C$

Note: In the above integration, while integrating $\int{\dfrac{1}{32\left( x-2 \right)}dx}-\int{\dfrac{1}{32\left( x+2 \right)}dx}$, you have got the result $\dfrac{1}{32}\ln \left| x-2 \right|-\dfrac{1}{32}\ln \left| x+2 \right|+C$ and in this result don’t forget to write the modulus sign because the expression inside the logarithm must be positive so make sure you won’t forget to write the modulus sign.