Question

Evaluate the integral $\int {\dfrac{1}{{x\left( {{x^3} + 8} \right)}}dx} $ .

Answer 1

Hint: Start by assuming the integral as I and try to simplify the denominator by taking out common terms, which would facilitate us to make a substitution with a new variable as there is no standard form of the obtained simplification. Convert all the variables in the substitution in a single variable, this can be done either by rearranging the terms or differentiating the substituting term. Solve the integral and substitute back the original value of the variable assumed, This would give us the correct solution.

Complete step-by-step answer:
Given, say $I = \int {\dfrac{1}{{x\left( {{x^3} + 8} \right)}}dx} $
Now . We will try to simplify the denominator by taking ${x^3}$ common , we get
$I = \int {\dfrac{1}{{x \cdot {x^3}\left( {1 + \dfrac{8}{{{x^3}}}} \right)}}dx} $
Now , we will make a substitution of $1 + \dfrac{8}{{{x^3}}}$ by t i.e. $1 + \dfrac{8}{{{x^3}}} = t$, We get
$I = \int {\dfrac{1}{{{x^4} \cdot t}}dx} \to eqn.1$
But in order to integrate all the variables must be in the form of t
So , Let us try to find relation between x and t , and for that we will differentiate
$1 + \dfrac{8}{{{x^3}}} = t$ with respect to x. we get
$\Rightarrow \dfrac{d}{{dx}}(1) + \dfrac{d}{{dx}}\left( {\dfrac{8}{{{x^3}}}} \right) = \dfrac{d}{{dx}}\left( t \right) $
Differentiation of constant term is always 0 and using the formula $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$, we will have
$
   \Rightarrow 0 + 8\dfrac{d}{{dx}}\left( {{x^{ - 3}}} \right) = \dfrac{d}{{dx}}\left( t \right) \\
   \Rightarrow 0 + \dfrac{{ - 24}}{{{x^4}}} = \dfrac{{dt}}{{dx}} \\
   \Rightarrow dx = - \dfrac{{{x^4}}}{{24}}dt \to eqn.2 \\
$
Now, Substituting the values in eqn.1 from eqn.2 , We get
$
  I = \int {\dfrac{1}{{{x^4} \cdot t}}\dfrac{{{x^4}}}{{( - 24)}}dt} \\
  I = - \dfrac{1}{{24}}\int {\dfrac{1}{t} \cdot dt} \\
  I = \dfrac{{ - 1}}{{24}}\int {\left( {\dfrac{{dt}}{t}} \right) = \dfrac{{ - 1}}{{24}}\ln \left( t \right) + c} \\
$
As , $\int {\dfrac{1}{x}dx} = \ln x + c$
Now , Substituting the value of t , we get
$I = \dfrac{{ - 1}}{{24}}\ln \left( {1 + \dfrac{8}{{{x^3}}}} \right) + c$

Note: Similar problems can be asked with different levels of intricacies, Therefore practice is recommended as such questions could be solved in many other possible ways but we must always go for the easiest alternative available and also according to the marks allotted to the question ( if provided in school exams). Students must also know all the standard forms and formulas related to indefinite integral.