Integral of 1 / x ^ 3

What you need:

• Integral rule for x ^ n

Simplify 1 / x ^ 3 - here's how to proceed

• Admittedly, the expression "1 / x ^ 3" is not easy to interpret, because behind it there is a (yet simple) broken rational function.
• First you form around f (x) = 1 / x ^ 3 = 1 / x³.
• Now you apply a power law, namely 1 / aⁿ = a^-n and you get: f (x) = x^-3.

Integral for functions with the negative power

• Just as one can find functions of the form f (x) = x^m with any Potencies m (here m can not only be a natural number, but also negative, a fraction or a real number) can be derived according to the known rule (with f (x) = x^m we have f '(x) = m _* x^m-1; where m can be any real number), you can also use the integral rule you are familiar with when integrating.
• Namely, ∫ x holds^m = 1 / (m + 1) _* x_m+1, whereby m does not necessarily have to be a natural number, with the exception of the case m = -1. The rule is easy to show by deriving (the reverse operation to integrate).
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Derive 2 by x - this is how it works with fractional-rational functions

If you want to derive the function "2 by x", you can do this with a little ...

• If you apply the rule, you can integrate any functions with any exponent (in your case also m = -3).
• You get: ∫ x^-3 = 1/(-3+1) _* x^-3+1 = = - 1/2 x^-2 = -1/2 _* 1 / x² = - 1 / (2x²), to show a few other notations, as well as in the somewhat more complicated notation -1/2 _* 1 / x ^ 2.

Conclusion: broken rational Functions of the type 1 / x ^ m can be integrated quite easily if you convert this into a function with a negative power and then apply the well-known integral rule. However, the procedure does not work for functions of the form 1 / (x² - 2x) or also 2x / (x + 1), since these are not simply broken functions. Other methods are necessary here, such as integration through substitution.