If x.f(x) = 3f2(x) +2 then ∫2x2−12x f(x) +f(x)(6f(x)−x) x2−f(x)2 dx

x.f1(x) +f(x) .1 =6 f(x)f1(x) ⇒f(x) =(6f(x)−x).f(x) ∫2x(x−6f(x)) +(6+f(x)−x).f1(x)(6f(x)−x)(x2−f(x))2 dx =−∫2x−f1(x)(x2 −f(x))2 dx =1x2−f(x) +c

If x.f(x) = 3f2(x) +2 then ∫2x2−12x f(x) +f(x)(6f(x)−x) x2−f(x)2 dx

x.f1(x) +f(x) .1 =6 f(x)f1(x) ⇒f(x) =(6f(x)−x).f(x) ∫2x(x−6f(x)) +(6+f(x)−x).f1(x)(6f(x)−x)(x2−f(x))2 dx =−∫2x−f1(x)(x2 −f(x))2 dx =1x2−f(x) +c

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If $x . f (x) = 3 f^{2} (x) + 2 t h e n \int \frac{2 x^{2} - 12 x f (x) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x$

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$\frac{2}{x^{2} + f (x)} + c$

$\frac{1}{x^{2} + f (x)} + c$

$\frac{1}{x^{2} - f (x)} + c$

$\frac{2}{x^{2} - f (x)} + c$

answer is D.

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Detailed Solution

$\begin{array}{l} x . f^{1} (x) + f (x) .1 = 6 f (x) f^{1} (x) \Rightarrow f (x) = (6 f (x) - x) . f (x) \\ \int \begin{array}{l} \frac{2 x (x - 6 f (x)) + (6 + f (x) - x) . f^{1} (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x \end{array} = - \int \frac{2 x - f^{1} (x)}{{(x^{2} - f (x))}^{2}} d x = \frac{1}{x^{2} - f (x)} + c \end{array}$