It is given that

$$6\int\limits_1^x {f(t)dt = 3xf(x) - {x^3}} - 5$$

$$ \Rightarrow 6f(x) = 3f(x) + 3xf'(x) - 3{x^2}$$

$$ \Rightarrow 3f(x) = 3xf'(x) - 3{x^2} \Rightarrow xf'(x) - f(x) = {x^2}$$

$$ \Rightarrow x{{dy} \over {dx}} - y = {x^2} \Rightarrow {{dy} \over {dx}} - {1 \over x}y = x$$ .... (1)

Now, $$I.F. = {e^{\int { - {1 \over x}dx} }} = {e^{ - {{\log }_e}x}}$$

Multiplying Eq. (1) both sides by $${1 \over x}$$, we get

$${1 \over x}{{dy} \over {dx}} - {1 \over {{x^2}}}y = 1 \Rightarrow {d \over {dx}}\left( {y.{1 \over x}} \right) = 1$$

Integrating, we get

$${y \over x} = x + c$$

Substituting x = 1 and y = 2, we get

$$ \Rightarrow 2 = 1 + c \Rightarrow c = 1 \Rightarrow y = {x^2} + x$$

$$ \Rightarrow f(x) = {x^2} + x \Rightarrow f(2) = 6$$