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Step-by-step explanation:

Note the expression below can be rewritten as

[tex]\dfrac{1}{(2x)^3} = \dfrac{1}{8x^3} = \dfrac{1}{8}x^{-3}[/tex]

Therefore, the we can rewrite the integral as

[tex]\displaystyle \int \dfrac{dx}{(2x)^3} = \frac{1}{8}\int x^{-3}dx[/tex]

[tex]\;\;\;\;\;\;\;\;\;\;= \dfrac{1}{8}\left(\dfrac{x^{-2}}{-2}\right) + C[/tex]

[tex]\;\;\;\;\;\;\;\;\;\;= -\dfrac{1}{16x^2} + C[/tex]

The indefinite integral of [tex]\int \frac{1}{(2x)^{3} }\, dx[/tex] is [tex]- \frac{1}{16x^{2} }+c[/tex].

What is the indefinite integral of [tex]\frac{1}{(2x)^{3} }[/tex]?

Given:

  • [tex]\int \frac{1}{(2x)^{3} }\, dx[/tex]

Find:

  • The indefinite integral of the given expression.

Solution:

[tex]\frac{1}{(2x)^{3} }[/tex]

We can also write the above expression as:

[tex]\frac{1}{(2x)^{3} } = \frac{1}{8x^{3} } = \frac{1}{8} x^{-3}[/tex]

Now, we solve it for indefinite integral, and we get;

[tex]\int \frac{dx}{(2x)^{3} } = \frac{1}{8} \int x^{-3} dx[/tex]

Now, applying the integration formula, we get;

[tex]= \frac{1}{8}\frac{x^{-2} }{-2} +c[/tex]

[tex]=-\frac{1}{16x^{2} } +c[/tex]

Hence, the indefinite integral of the  [tex]\int \frac{1}{(2x)^{3} }\, dx[/tex] is [tex]- \frac{1}{16x^{2} }+c[/tex].

To learn more about indefinite integral, refer to:

https://brainly.com/question/12231722

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