Question

How do you integrate $f\left( x \right) = \left( {{x^2} + x + 1} \right)\left( {x - 1} \right)$ using the product rule?

Answer 1

There is no product rule for integration but there is an integration method called integration by parts that we closely related to the product rule. So, in order to solve this problem, we will use integration by parts methods. For this we will let $u = \left( {x - 1} \right)$ and $v = \left( {{x^2} + x + 1} \right)$ .Then we will substitute the values in the formula i.e., $\int {\left( {uv} \right)} {\text{ }}dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right){\text{ }}} } dx$ .After that we will use the power rule i.e., $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}$ for further simplification. Hence, we will get the required result.

Complete step by step answer:
We have given the function as
$f\left( x \right) = \left( {{x^2} + x + 1} \right)\left( {x - 1} \right)$
Now we know that the integration by parts formula is:
$\int {\left( {uv} \right)} {\text{ }}dx = u\int {vdx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right){\text{ }}} } dx$

Let us assume,
$u = \left( {x - 1} \right)$ and $v = \left( {{x^2} + x + 1} \right)$
Therefore, on substituting the values in the formula, we get
$\int {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} {\text{ }}dx = \left( {x - 1} \right)\int {\left( {{x^2} + x + 1} \right)dx - \int {\left( {\dfrac{{d\left( {x - 1} \right)}}{{dx}}\int {\left( {{x^2} + x + 1} \right)dx} } \right){\text{ }}} } dx$
Now we know that
$\dfrac{{d\left( {x - a} \right)}}{{dx}} = 1$
And according to the power rule of integration
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}$

Therefore, we have
$\int {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} {\text{ }}dx = \left( {x - 1} \right)\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x} \right) - \int {\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x} \right)} dx$
Again, integrating using power rule, we get
$\Rightarrow \int {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)} {\text{ }}dx = \left( {x - 1} \right)\left( {\dfrac{{{x^3}}}{3} + \dfrac{{{x^2}}}{2} + x} \right) - \left( {\dfrac{{{x^4}}}{{12}} + \dfrac{{{x^3}}}{6} + \dfrac{{{x^2}}}{2}} \right) + c$
Now on multiplying, we get
$\Rightarrow \int {\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)} {\text{ }}dx = \left( {\dfrac{{{x^4}}}{3} + \dfrac{{{x^3}}}{2} + {x^2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^2}}}{2} - x} \right) - \left( {\dfrac{{{x^4}}}{{12}} + \dfrac{{{x^3}}}{6} + \dfrac{{{x^2}}}{2}} \right) + c$

On combining the like terms, we get
$\Rightarrow \int {\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)} {\text{ }}dx = \left( {\dfrac{{{x^4}}}{3} - \dfrac{{{x^4}}}{{12}}} \right) + \left( {\dfrac{{{x^3}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^3}}}{6}} \right) + \left( {{x^2} - \dfrac{{{x^2}}}{2} - \dfrac{{{x^2}}}{2}} \right) + \left( { - x} \right) + c$
On subtracting, we get
$\therefore \int {\left( {x - 1} \right)\left( {{x^2} + x - 1} \right)} {\text{ }}dx = \dfrac{{{x^4}}}{4} - x + c$

Hence, the integration of $f\left( x \right) = \left( {{x^2} + x + 1} \right)\left( {x - 1} \right)$ is $\dfrac{{{x^4}}}{4} - x + c$.

Note: Integration by parts method is used for integrating the product of two functions. This method is used for finding the integrals by reducing them into the standard form. The formula of integration by parts method can also be given as follows:
$\int {udv = uv - \int {vdu} }$
In general, this question can be solved by just multiplying the terms and then by integrating each term using the power rule. But in the question, it was given that we have to use product rules, so we used that method. We can also compare the above answer with the answer that we get, if we just perform the multiplication:As we have given,
$f\left( x \right) = \left( {{x^2} + x + 1} \right)\left( {x - 1} \right)$
On multiplying, we get
$f\left( x \right) = \left( {{x^3} + {x^2} + x - {x^2} - x - 1} \right)$
On simplifying, we get
$\Rightarrow f\left( x \right) = {x^3} - 1$
Now on integrating using power rule, we get
$\int {f\left( x \right)} {\text{ }}dx = \int {\left( {{x^3} - 1} \right){\text{ }}} dx$
$\Rightarrow \int {f\left( x \right)} {\text{ }}dx = \dfrac{{{x^4}}}{4} - x + c$
which is the required answer.
Hence, we get the correct answer.