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Question

Question: How do you integrate \[(\dfrac{1}{{{x^4}}})dx\]?...

How do you integrate (1x4)dx(\dfrac{1}{{{x^4}}})dx?

Explanation

Solution

we have to use normal integration formulas to solve this problem. The variable is in the denominator so we have to first make it a numerator and then use integration formulas to solve it. Integrating a function with respect to x is just finding the area covered by the graph of that function and the x-axis. Integration is just the inverse of differentiation. So if we just differentiate the answer we will get our question.

Complete Step by Step Solution:
According to the question, we have to integrate the function (1x4)dx(\dfrac{1}{{{x^4}}})dx,
So we know the formula of integration as xndx=xn+1n+1+c\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c}
Where n is the power of the variable x and c is the constant term we always add after every indefinite integration
In this case, we have the value of n=4n = - 4, because we can write the given question in the form x4dx\int {{x^{ - 4}}} dx
Now, we have to simplify it
x4dx=x4+14+1+c\Rightarrow \int {{x^{ - 4}}} dx = \dfrac{{{x^{ - 4 + 1}}}}{{ - 4 + 1}} + c
x4+14+1=x33+c\Rightarrow \dfrac{{{x^{ - 4 + 1}}}}{{ - 4 + 1}} = \dfrac{{{x^{ - 3}}}}{{ - 3}} + c

x33=13x3+c \Rightarrow \dfrac{{{x^{ - 3}}}}{{ - 3}} = \dfrac{{ - 1}}{{3{x^3}}} + c
Hence, this is our answer.

Note: We can observe that while doing the integration we have to bring the variable in the denominator to the numerator for simple calculation. We should also be aware that every time we do any indefinite integration we have to add a constant of integration which is c in this case. It is done because when we differentiate our answer, we will get our question but it may be that there might be a constant which will turn into zero. We have to remember all the basic integration formulas for ease while solving any type of integration problem in the future. Multiplying a negative one to the power of any variable or constant will just convert it into its own multiplicative inverse.