Question

How do you integrate\[\int{\dfrac{{{x}^{2}}}{1-{{x}^{3}}}dx}\] ?

Answer 1

Hint:In the given question, we have been asked to integrate the following function. In order to solve the question, we integrate the numerical by following the substitution method. We substitute \[u=1-{{x}^{3}}\]and after integrating the given function we undo the substitution 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 given,
\[\Rightarrow \int{\dfrac{{{x}^{2}}}{1-{{x}^{3}}}dx}\]
Substitute \[u=1-{{x}^{3}}\] in the above equation,
\[\Rightarrow u=1-{{x}^{3}}\]
Differentiate ‘u’ with respect to ‘x’, we get
\[\Rightarrow \dfrac{du}{dx}=-3{{x}^{2}}\]
Solving the above for the value of dx, we get
\[\Rightarrow dx=-\dfrac{1}{3}du\]
Substituting \[dx=-\dfrac{1}{3}du\] and\[u=1-{{x}^{3}}\], we obtain
\[\Rightarrow \dfrac{1}{3}\int{\dfrac{1}{u}du}\]
Now, solving
\[\Rightarrow \int{\dfrac{1}{u}du}\]
This is the standard integral, which is equal to \[\ln \left( u \right)\]
\[\Rightarrow\int{\dfrac{1}{u}du=\ln \left( u \right)}\]
Now taking,
\[\Rightarrow \dfrac{1}{3}\int{\dfrac{1}{u}du}\]
Combining the previously solved integrals, we get
\[\Rightarrow -\dfrac{\ln \left( u \right)}{3}\]
Replacing the substitution\[u=1-{{x}^{3}}\], we get
\[\Rightarrow -\dfrac{\ln \left( 1-{{x}^{3}} \right)}{3}\]
\[\therefore \int{\dfrac{{{x}^{2}}}{1-{{x}^{3}}}dx}=\]\[-\dfrac{\ln \left( 1-{{x}^{3}} \right)}{3}\]
Hence, it is the required integration.

Note:Integration is the part of calculus that includes the differentiation. Integration refers to the, add up smaller parts of any area given, volume given, etc to represents the whole value. To solve any given numerical, or function, there are different types of integration methods like integration by substitution, integration by parts, etc. We should remember the property or the formulas of integration, this would make it easier to solve the question. There are also many other methods for integration. These are integration by substitution and integration by partial fractions. You should always remember all the methods for integration so that we can easily choose which method is suitable for solving the particular type of question. We should do all the calculations carefully and explicitly to avoid making errors.