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

I=∫2x(x−6f(x))+f(x)(6f(x)−x)(x2−f(x))2dx= −∫2x(x2−f(x))2dx+∫f(x)(6f(x)−x)(x2−f(x))2∫f1(x)−2x(x2−f(x1))2dx − ∫f1(x)(x2−f(x))2+ ∫f(x)(6f(x)−x)(x2−f(x))2dx=∫f1(x)−2x(x2−f(x))2 dx +∫(6f(x)−x)(−f1(x))+f(x)(6f(x)−x)(x2−f(x))2 dx=∫f1(x)−2x(x2−f(x))2 dx+ ∫−f(x)+f(x)(6f(x)−x)(x2−f(x))2dx∫−dtt2 |(2x−f1(x)dx=+dtx2−f(x)=t| xf(x)+f(x)=6f(x)f1(x)x f(x) = 3(f(x))2+2= 1t+C 1x2−f(x)+C

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

I=∫2x(x−6f(x))+f(x)(6f(x)−x)(x2−f(x))2dx= −∫2x(x2−f(x))2dx+∫f(x)(6f(x)−x)(x2−f(x))2∫f1(x)−2x(x2−f(x1))2dx − ∫f1(x)(x2−f(x))2+ ∫f(x)(6f(x)−x)(x2−f(x))2dx=∫f1(x)−2x(x2−f(x))2 dx +∫(6f(x)−x)(−f1(x))+f(x)(6f(x)−x)(x2−f(x))2 dx=∫f1(x)−2x(x2−f(x))2 dx+ ∫−f(x)+f(x)(6f(x)−x)(x2−f(x))2dx∫−dtt2 |(2x−f1(x)dx=+dtx2−f(x)=t| xf(x)+f(x)=6f(x)f1(x)x f(x) = 3(f(x))2+2= 1t+C 1x2−f(x)+C

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If $x . f (x) = 3 {(f (x))}^{2} + 2$ then $\int \frac{2 x^{2} - 12 x f (x) + f (x)}{(6 f (x) - x) (x^{2} - {(f (x))}^{2}} d x$ is equal to

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$\frac{1}{{(6 f (x) - x)}^{2}} + C$

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

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

$\frac{1}{(6 f (x) - x^{2})} + C$

answer is B.

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Detailed Solution

$I = \int \frac{2 x (x - 6 f (x)) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x = - \int \frac{2 x}{{(x^{2} - f (x))}^{2}} d x + \int \frac{f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}}$

$\int \frac{f^{1} (x) - 2 x}{{(x^{2} - f (x_{1}))}^{2}} d x - \int \frac{f^{1} (x)}{{(x^{2} - f (x))}^{2}} + \int \frac{f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x$

$= \int \frac{f^{1} (x) - 2 x}{{(x^{2} - f (x))}^{2}} d x + \int \frac{(6 f (x) - x) (- f^{1} (x)) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x$

$= \int \frac{f^{1} (x) - 2 x}{{(x^{2} - f (x))}^{2}} d x + \int \frac{- f (x) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x$

$\int \frac{- d t}{t^{2}} | \overset{x^{2} - f (x) = t}{(2 x - f^{1} (x) d x = + d t} | \overset{x f (x) = 3 {(f (x))}^{2} + 2}{x f (x) + f (x) = 6 f (x) f^{1} (x)}$