If ∫ f(x)dx = 3[f(x)]2 + c (where c is the constant of integration)...

Differentiating the given integral with respect to $‘x’,$ we get,
$f$\left(x\right) = 6f \left(x\right) ⋅ f^{′} \left(x\right)$or$\dfrac{1}{6} =f^{′} $\left(x\right)<br />On integrating, we get,<br />$∫ $\dfrac{1}{6} dx = ∫ f^{′} \left(x\right) dx$
$⇒\dfrac{x}{6}=$f\left(x\right)$+$ c$<br />As$f$\left(1\right) = \dfrac{1}{6}$⇒ c = 0$<br />$⇒ f \left(x\right)=\dfrac{x}{6}$<br />$⇒ f \left(6\pi\right) =\pi