Question

Find the integral $\int{\dfrac{x}{{{x}^{4}}-1}dx}$?

Answer 1

Hint: Assume the value of the given integral as ‘I’. Write ${{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}}$ and substitute ${{x}^{2}}=k$. Now, differentiate this substituted value and find the value of $xdx$ in terms of $dk$ and substitute it in the integral. Use the partial fraction to simplify the function in the integral and use the formula $\int{\dfrac{1}{ak+b}dk}=\dfrac{1}{a}\ln \left( ak+b \right)$ to get the required integral. Here, ‘a’ and ‘b’ are constants. Finally substitute back the value of k in terms of x and add the constant of indefinite integration ‘C’ at last.

Complete step by step solution:
Here we have been provided with the function $\dfrac{x}{{{x}^{4}}-1}$ and we are asked to integrate it. Let us assume the integral as I, so we have,
$\Rightarrow I=\int{\dfrac{x}{{{x}^{4}}-1}dx}$
Now in the denominator we have the exponential term ${{x}^{4}}$ which can be written as \[{{\left( {{x}^{2}} \right)}^{2}}\] using the formula: ${{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}$. So we get the integral,
$\Rightarrow I=\int{\dfrac{x}{{{\left( {{x}^{2}} \right)}^{2}}-1}dx}$
Let us use the substitution method to solve this integral, so substituting ${{x}^{2}}=k$ and differentiating both the sides to find the value of $xdx$ in terms of $dk$ we get,
$\begin{align}
  & \Rightarrow d\left( {{x}^{2}} \right)=dk \\
 & \Rightarrow 2xdx=dk \\
 & \Rightarrow xdx=\dfrac{dk}{2} \\
\end{align}$
Substituting the assumed and above obtained relation in the integral I we get,
\[\begin{align}
  & \Rightarrow I=\int{\left( \dfrac{1}{{{k}^{2}}-1} \right)\dfrac{dk}{2}} \\
 & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{1}{{{k}^{2}}-1}dk} \\
\end{align}\]
Using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ we get,
\[\begin{align}
  & \Rightarrow I=\int{\left( \dfrac{1}{{{k}^{2}}-1} \right)\dfrac{dk}{2}} \\
 & \Rightarrow I=\dfrac{1}{2}\int{\dfrac{1}{\left( k+1 \right)\left( k-1 \right)}dk} \\
\end{align}\]
We can write the numerator of the function inside the integral as $1=\left( k+1 \right)-\left( k-1 \right)$ so we get,
\[\Rightarrow I=\dfrac{1}{2}\int{\dfrac{\left( k+1 \right)-\left( k-1 \right)}{\left( k+1 \right)\left( k-1 \right)}dk}\]
Breaking the terms we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{1}{2}\int{\left[ \dfrac{1}{\left( k-1 \right)}-\dfrac{1}{\left( k+1 \right)} \right]dk} \\
 & \Rightarrow I=\dfrac{1}{2}\left[ \int{\dfrac{dk}{\left( k-1 \right)}}-\int{\dfrac{dk}{\left( k+1 \right)}} \right] \\
\end{align}\]
The above integral is of the form $\int{\dfrac{1}{ak+b}dk}$ where ‘a’ and ‘b’ are constants. Its solution is $\dfrac{1}{a}\ln \left( ak+b \right)$, so we get our integral as:
\[\Rightarrow I=\dfrac{1}{2}\left[ \ln \left( k-1 \right)-\ln \left( k+1 \right) \right]\]
Using the property of log given as $\log m-\log n=\log \left( \dfrac{m}{n} \right)$ and substituting back the value of k in terms of x we get,
\[\Rightarrow I=\dfrac{1}{2}\left[ \ln \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right) \right]\]
Now, since the given integral was an indefinite integral and therefore we need to add a constant of integration (C) in the expression obtained for I. So we get,
\[\therefore I=\dfrac{1}{2}\left[ \ln \left( \dfrac{{{x}^{2}}-1}{{{x}^{2}}+1} \right) \right]+C\]
Hence, the above relation is our answer.

Note: Note that the substitution method is an important method to solve integrals. Here, you may see that we have applied the method of partial fraction to break the terms so you must find a way to simplify such situations otherwise it will be difficult to integrate the function. Remember the integral formulas of all the common functions. At last, do not forget to add the constant of integration (C) as we are finding indefinite integral and not definite integral.