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Question: What is the antiderivative of \(\dfrac{4}{{{x}^{2}}}\)?...

What is the antiderivative of 4x2\dfrac{4}{{{x}^{2}}}?

Explanation

Solution

We solve this problem by the integral of the given function because the anti – derivative is the second name of integration. We use the standard power rule of integration that is given as,
xn.dx=xn+1n+1+C\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+C
Convert the given function as in the above formula and apply the formula to get the required answer.

Complete step-by-step answer:
We are asked to find the anti – derivative of 4x2\dfrac{4}{{{x}^{2}}}
Let us assume that the given function as,
f(x)=4x2\Rightarrow f\left( x \right)=\dfrac{4}{{{x}^{2}}}
We know that anti – derivative is nothing but the integration of the function.
Now, let us integrate the given function with respect to x'x' then we get,
f(x).dx=4x2dx\Rightarrow \int{f\left( x \right).dx}=\int{\dfrac{4}{{{x}^{2}}}dx}
We know that in integration the constant can be taken out without any change in integration.
By taking the constant 4 outside the integral then we get,
f(x).dx=41x2.dx\Rightarrow \int{f\left( x \right).dx}=4\int{\dfrac{1}{{{x}^{2}}}.dx}
We know that if a term having the power in denominator then it can be taken to numerator by making the power to negative that is, 1an=an\dfrac{1}{{{a}^{n}}}={{a}^{-n}}
By using this rule of exponents in the above integral then we get,
f(x).dx=4x2.dx\Rightarrow \int{f\left( x \right).dx}=4\int{{{x}^{-2}}.dx}
Now, let us apply the power rule on integration which is given as,
xn.dx=xn+1n+1+C\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+C
By applying the power rule to above integral then, we get,
f(x).dx=4(x2+12+1)+C f(x).dx=4x1+C \begin{aligned} & \Rightarrow \int{f\left( x \right).dx}=4\left( \dfrac{{{x}^{-2+1}}}{-2+1} \right)+C \\\ & \Rightarrow \int{f\left( x \right).dx}=-4{{x}^{-1}}+C \\\ \end{aligned}
Now, by using the exponent rule we used above in reverse order in above equation then we get,
f(x).dx=4x+C\Rightarrow \int{f\left( x \right).dx}=\dfrac{-4}{x}+C
Therefore, we can conclude that the anti – derivative of 4x2\dfrac{4}{{{x}^{2}}} is given as,
4x2.dx=4x+C\therefore \int{\dfrac{4}{{{x}^{2}}}.dx}=\dfrac{-4}{x}+C

Note: We need to note that the power rule for integration is applied when the function is in numerator either with positive power or negative power. So, the power rule is given as,
xn.dx=xn+1n+1+C\int{{{x}^{n}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+C
Sometimes, students may directly apply the power rule even if the function is in denominator like,
1xn.dx=xn+1n+1+C\int{\dfrac{1}{{{x}^{n}}}.dx}=\dfrac{{{x}^{n+1}}}{n+1}+C
This is a very wrong thing to do. Also even if the value of n'n' is negative the power is applied.