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Question: What is the antiderivative of \(\dfrac{1}{{{x^3}}}\) ?...

What is the antiderivative of 1x3\dfrac{1}{{{x^3}}} ?

Explanation

Solution

In calculus, anti derivative of a function is same as indefinite integral of the given function. The given question requires us to integrate a function of x with respect to x. Integration gives us a family of curves. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. integral is always found with respect to some variable, which in this case is x.

Complete step by step solution:
The given question requires us to integrate a rational function 1x3\dfrac{1}{{{x^3}}} in variable x whose numerator is 11 and denominator is x3{x^3}. So, we first represent the function in negative power form and then integrate the function directly using the power rule of integration.
So, we can write 1x3\dfrac{1}{{{x^3}}} as x3{x^{ - 3}}.
Hence, we have to integrate x3{x^{ - 3}} with respect to x.
So, we have to evaluate x3dx\int {{x^{ - 3}}} dx.
Now, we know the power rule of integration. According to the power rule of integration, the integral of xn{x^n} with respect to x is x(n+1)n+1+C\dfrac{{{x^{\left( {n + 1} \right)}}}}{{n + 1}}+C.
So, we get, x3dx=x3+13+1+C\int {{x^{ - 3}}} dx = \dfrac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}+C
x3dx=x22+C\Rightarrow \int {{x^{ - 3}}} dx = \dfrac{{{x^{ - 2}}}}{{ - 2}}+C
Simplifying the expression further, we get,
x3dx=12x2+C\Rightarrow \int {{x^{ - 3}}} dx = \dfrac{1}{{ - 2{x^2}}}+C
So, (12x2)+C\left( {\dfrac{1}{{ - 2{x^2}}}} \right)+C is the antiderivative of the given function 1x3\dfrac{1}{{{x^3}}}.

Note:
The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the constant.