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].
Given:
Find:
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:
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