Respuesta :

Answer: None of the above

Step-by-step explanation:

Given

[tex]\frac{\mathrm{d} \bar{c}}{\mathrm{d} x}=-6x^{-2}+\frac{1}{6}[/tex]

[tex]d\bar{c}=\left ( -6x^{-2}+\frac{1}{6}\right )dx[/tex]

Now, integrating both sides

[tex]\int d\bar{c}=\int \left ( -6x^{-2}+\frac{1}{6}\right )dx[/tex]

[tex]\bar{c}=-6\frac{x^{-2+1}}{-2+1}+\frac{1}{6}x+k_1[/tex]

[tex]\bar{c}=6x^{-1}+\frac{1}{6}x+k_1[/tex]

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