How do you solve\[\int{\dfrac{x}{{{x}^{4}}-1}dx}\]?

Hint: In the given question, we have been asked to integrate the following numerical. In order to solve the question, we integrate the numerical by following the substitution method. We replace \[u=\dfrac{{{x}^{2}}}{2}\]and replace \[{{x}^{4}}=4{{u}^{2}}\]and solve the numerical by further integration. After we have simplified our sum, we just need to integrate the terms and then we replace the substituted variables by the original variables.

Complete step-by-step answer:
We have the given function,
\[\Rightarrow \int{\dfrac{x}{{{x}^{4}}-1}dx}\]
Substitute \[u=\dfrac{{{x}^{2}}}{2}\], and differentiate with it respect to \[x\], we get
\[\Rightarrow u=\dfrac{{{x}^{2}}}{2}\]
Differentiate the above equation,
\[\Rightarrow \dfrac{du}{dx}=x\]
Simplifying the above, we get
\[\Rightarrow dx=\dfrac{1}{x}du\]
Use, \[{{x}^{4}}=4{{u}^{2}}\],
As \[u=\dfrac{{{x}^{2}}}{2}\]then \[4{{u}^{2}}=4{{\left( \dfrac{{{x}^{2}}}{2} \right)}^{2}}=4\left( \dfrac{{{x}^{4}}}{4} \right)={{x}^{4}}\]
Substitute, \[{{x}^{4}}=4{{u}^{2}}\] and \[dx=\dfrac{1}{x}du\]in the given function in the question, we get
\[\Rightarrow \int{\dfrac{x}{4{{u}^{2}}-1}\times \dfrac{1}{x}du}=\int{\dfrac{1}{4{{u}^{2}}-1}du}\]
Factoring the denominator by using the identity i.e. \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we get
\[\Rightarrow \int{\dfrac{1}{\left( 2u-1 \right)\left( 2u+1 \right)}du}\]
Perform partial fraction decomposition,
\[\Rightarrow \int{\left( \dfrac{1}{2\left( 2u-1 \right)}-\dfrac{1}{2\left( 2u+1 \right)}du \right)}\]
Applying the linearity, we get
\[\Rightarrow \dfrac{1}{2}\int{\dfrac{1}{2u-1}du}-\dfrac{1}{2}\int{\dfrac{1}{2u+1}du}\]
\[\Rightarrow \]Now solving,
\[\Rightarrow \int{\dfrac{1}{2u-1}du}\]
Substitute \[v=2u-1\]
Differentiate the following, we get
\[\Rightarrow \dfrac{dv}{du}=2\]
\[\Rightarrow du=\dfrac{1}{2}dv=\dfrac{1}{2}\int{\dfrac{1}{v}dv}\]
Now solving,
\[\Rightarrow \int{\dfrac{1}{v}dv}\]
This is the standard integral which is equal to \[\ln \left( v \right)\]
Now, substitute this value in previously solved integral, we get
\[\Rightarrow \dfrac{1}{2}\int{\dfrac{1}{v}dv}=\dfrac{\ln \left( v \right)}{2}\]
Now replacing \[v\] by \[2u-1\], we get
\[\Rightarrow \dfrac{\ln \left( 2u-1 \right)}{2}\]
\[\Rightarrow \]Now solving,
\[\Rightarrow \int{\dfrac{1}{2u+1}du}\]
Substitute \[v=2u+1\]
Differentiate the following, we get
\[\Rightarrow \dfrac{dv}{du}=2\]
\[\Rightarrow du=\dfrac{1}{2}dv=\dfrac{1}{2}\int{\dfrac{1}{v}dv}\]
Now solving,
\[\Rightarrow \int{\dfrac{1}{v}dv}\]
This is the standard integral which is equal to \[\ln \left( v \right)\]
Now, substitute this value in previously solved integral, we get
\[\Rightarrow \dfrac{1}{2}\int{\dfrac{1}{v}dv}=\dfrac{\ln \left( v \right)}{2}\]
Now replacing \[v\] by \[2u+1\], we get
\[\Rightarrow \dfrac{\ln \left( 2u+1 \right)}{2}\]
Now, plugged in solved integrals, we obtain
\[\Rightarrow \dfrac{1}{2}\int{\dfrac{1}{2u-1}du}-\dfrac{1}{2}\int{\dfrac{1}{2u+1}du}\]
\[\Rightarrow \dfrac{\ln \left( 2u-1 \right)}{4}-\dfrac{\ln \left( 2u+1 \right)}{4}\]
Replacing the value of \[u=\dfrac{{{x}^{2}}}{2}\], we get
\[\Rightarrow \dfrac{\ln \left( 2\times \dfrac{{{x}^{2}}}{2}-1 \right)}{4}-\dfrac{\ln \left( 2\times \dfrac{{{x}^{2}}}{2}+1 \right)}{4}\]
\[\Rightarrow \dfrac{\ln \left( {{x}^{2}}-1 \right)}{4}-\dfrac{\ln \left( {{x}^{2}}+1 \right)}{4}\]
Applying the absolute value function, we get
\[\Rightarrow \int{\dfrac{x}{{{x}^{4}}-1}dx}=\dfrac{\ln \left( \left| {{x}^{2}}-1 \right| \right)}{4}-\dfrac{\ln \left( {{x}^{2}}+1 \right)}{4}+C\]
\[\Rightarrow \dfrac{\ln \left( \left| {{x}^{2}}-1 \right| \right)-\ln \left( {{x}^{2}}+1 \right)}{4}+C\], it is the required solution.

Note: Here, we need to remember that we have to put the constant term C after the integration equation and the value of the given constant i.e. C, it can be any value 0 equal to zero also. In order to solve the question that is given above, students need to know the basic formula of integration and they should very well keep all the standard integral into their mind because sometimes the given integration is the standard integral and we do not need to solve the question further and directly write the resultant integral.