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

xf(x)=3f2(x)+2 Differentiating both sides wrtx xf′(x)+f(x)=6f(x)f′(x)xf′(x)−6f(x)f(x)=−f(x)x−6f(x)f′(x)=−f(x)f′(x)=−f(x)x−6f(x)f′(x)=f(x)6f(x)−x----1 Now =∫2x2−12xf(x)+f(x)(6f(x)−x)x2−f(x)2dx=∫2x(x−6f(x))+f(x)(6f(x)−x)x2−f(x)2dx=∫−2xx2−f(x)2dx+∫f(x)f(x)f(x)x2−f(x)2dx=∫−2xx2−f(x)2dx+∫f′(x)x2−f(x)2dx=∫−2x−f′(x)x2−f(x)2dx=∫1x2−f(x)+C Put x2−f(x)=t2x−f′(x)dx=dt=−∫1t2dt=1t+C=1x2−f(x)+C

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

xf(x)=3f2(x)+2 Differentiating both sides wrtx xf′(x)+f(x)=6f(x)f′(x)xf′(x)−6f(x)f(x)=−f(x)x−6f(x)f′(x)=−f(x)f′(x)=−f(x)x−6f(x)f′(x)=f(x)6f(x)−x----1 Now =∫2x2−12xf(x)+f(x)(6f(x)−x)x2−f(x)2dx=∫2x(x−6f(x))+f(x)(6f(x)−x)x2−f(x)2dx=∫−2xx2−f(x)2dx+∫f(x)f(x)f(x)x2−f(x)2dx=∫−2xx2−f(x)2dx+∫f′(x)x2−f(x)2dx=∫−2x−f′(x)x2−f(x)2dx=∫1x2−f(x)+C Put x2−f(x)=t2x−f′(x)dx=dt=−∫1t2dt=1t+C=1x2−f(x)+C

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$If x f (x) = 3 f^{2} (x) + 2 Then \int \frac{2 x^{2} - 12 x f (x) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x =$

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

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

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

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

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Detailed Solution

$x f (x) = 3 f^{2} (x) + 2$

$\begin{array}{l} Differentiating both sides wrtx \\ x f^{'} (x) + f (x) = 6 f (x) f^{'} (x) \\ x f^{'} (x) - 6 f (x) f (x) = - f (x) \\ (x - 6 f (x) f^{'} (x) = - f (x) \\ f^{'} (x) = \frac{- f (x)}{x - 6 f (x)} \\ f^{'} (x) = \frac{f (x)}{6 f (x) - x} - - - - (1) \end{array}$

$\begin{aligned} Now = \int \frac{2 x^{2} - 12 x f (x) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x \\ = \int \frac{2 x (x - 6 f (x)) + f (x)}{(6 f (x) - x) {(x^{2} - f (x))}^{2}} d x \end{aligned}$

$\begin{array}{l} = \int \frac{- 2 x}{{(x^{2} - f (x))}^{2}} d x + \int \frac{f (x)}{\frac{f (x)}{f^{(x)}} {(x^{2} - f (x))}^{2}} d x \\ = \int \frac{- 2 x}{{(x^{2} - f (x))}^{2}} d x + \int \frac{f^{'} (x)}{{(x^{2} - f (x))}^{2}} d x \\ = \int \frac{- 2 x - f^{'} (x)}{{[x^{2} - f (x)]}^{2}} d x \\ = \int \frac{1}{x^{2} - f (x)} + C \\ Put x^{2} - f (x) = t \\ (2 x - f^{'} (x) d x = d t) \\ = - \int \frac{1}{t^{2}} d t = \frac{1}{t} + C \\ = \frac{1}{x^{2} - f (x)} + C \end{array}$