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Question: How do you integrate \(\int {\left( {4{x^3} - \dfrac{1}{{{x^2}}}} \right)dx} \)?...

How do you integrate ∫(4x3−1x2)dx\int {\left( {4{x^3} - \dfrac{1}{{{x^2}}}} \right)dx} ?

Explanation

Solution

We will first write the general integration of xn{x^n} and then just put in the values of n = 3 and n = - 2 for finding both the integration and sum them up to find the answer.

Complete step-by-step answer:
We are given that we are required to find the value of ∫(4x3−1x2)dx\int {\left( {4{x^3} - \dfrac{1}{{{x^2}}}} \right)dx} .
This can be written as: ∫4x3dx−∫1x2dx\int {4{x^3}dx} - \int {\dfrac{1}{{{x^2}}}dx} .
This can be further written as ∫4x3dx−∫x−2dx\int {4{x^3}dx} - \int {{x^{ - 2}}dx} …………(1)
We see that the integration of xn{x^n} is given by: ∫xndx=xn+1n+1\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}
Now, let us for once put n = 3 in the above formula to get the following:-
⇒∫x3dx=x44\Rightarrow \int {{x^3}dx} = \dfrac{{{x^4}}}{4} ……………(2)
Now, let us for once put n = - 2 in the above formula to get the following:-
⇒∫x−2dx=x−1−1\Rightarrow \int {{x^{ - 2}}dx} = \dfrac{{{x^{ - 1}}}}{{ - 1}} ……………(3)
Putting the values we got from (2) and (3) in equation (1), we will then obtain:-
⇒∫4x3dx−∫x−2dx=x44+1x\Rightarrow \int {4{x^3}dx} - \int {{x^{ - 2}}dx} = \dfrac{{{x^4}}}{4} + \dfrac{1}{x}

Since, this is an indefinite integral, we will have to introduce a constant c in it. So, we finally have the answer as: ∫(4x3−1x2)dx=x44+1x\int {\left( {4{x^3} - \dfrac{1}{{{x^2}}}} \right)dx} = \dfrac{{{x^4}}}{4} + \dfrac{1}{x}.

Note:
The students must note that in the starting also we have used a theorem which helped us to divide the given integral into two parts. The theorem states that: for any two real valued functions f(x) and g(x), we have: ∫(f+g)(x)dx=∫f(x)dx+∫g(x)dx\int {(f + g)(x)dx = \int {f(x)dx} } + \int {g(x)dx} . This theorem is true for a finite number of functions instead of 2. So, even if you are given ‘n’ numbers of functions, you may use this as well.
The students must also note that we always introduce a constant c in the indefinite integral because we are not given the limits to it and it may take some other value. Let us look at an example to see that as well.
If we are given that dydx=4\dfrac{{dy}}{{dx}} = 4, then its integration will yield: y = 4x + C.
This C is introduced here because the differential of 4x + 2 and 4x + 3 or anything will be the same that is 4. So, we are not very sure of the function and we introduce a constant.