Question

Evaluate the following expression $\int{\dfrac{dx}{x\left( {{x}^{3}}+8 \right)}}$.

Answer 1

Hint: Whenever we have a complex function, it is always easier to replace the function with some other simpler function so as to make the integration easier. In this question, we are going to replace the polynomial in the denominator with a simpler function and then integrate by substitution. We will also use the standard integral that $\int{\dfrac{dx}{x}=\log \left( x \right)+C}$.

Complete step-by-step solution
Let us rewrite the given integral in the following way,
$I=\int{\dfrac{dx}{x\left( {{x}^{3}}+8 \right)}=}\dfrac{dx}{{{x}^{4}}+8x}$
Now, on taking ${{x}^{4}}$ common from both the terms in the denominator, we will get that,
\[I=\int{\dfrac{dx}{{{x}^{4}}\left( 1+\dfrac{8}{{{x}^{3}}} \right)}.........}(i)\]
Now, let us consider $u=1+\dfrac{8}{{{x}^{3}}}$. On differentiating both the sides of $u=1+\dfrac{8}{{{x}^{3}}}$, we will get that $\dfrac{dx}{{{x}^{4}}}=-\dfrac{du}{24}$.
On replacing the values of the integral in equation (i) with the above-mentioned values, we will get as follows.
$I=\int{-\dfrac{1}{24}\dfrac{du}{u}}.........(ii)$
We can see that this is a standard integral, $\int{\dfrac{dx}{x}}$. And we already know that, $\int{\dfrac{dx}{x}=\log \left( x \right)+C}$.
So, on using this property in equation (ii), we will get that,
$I=-\dfrac{1}{24}\log \left( u \right)+C$
Now, on substituting back the value of $u=1+\dfrac{8}{{{x}^{3}}}$ in the above equation, we will get as follows.
$I=-\dfrac{1}{24}\log \left( 1+\dfrac{8}{{{x}^{3}}} \right)+C$
On simplifying it, we will get,
$I=-\dfrac{1}{24}\log \left( \dfrac{{{x}^{3}}+8}{{{x}^{3}}} \right)+C$
Therefore, we get the answer as, $-\dfrac{1}{24}\log \left( \dfrac{{{x}^{3}}+8}{{{x}^{3}}} \right)+C$

Note: While substituting the functions, one should always be careful while changing the differentials. And one should not forget the constant of integration as this is indefinite integration after all. Sometimes, the students forget the term $-\dfrac{1}{24}$ while changing the variables which might give you the wrong value of the integral.