If 5(f(x))2=xf(x)+30 and ∫3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2dx=ABx3+Df(x)+C then A+B+D=

∫3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2dx=ABx3+Df(x)+C⇒3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2=−ABx3+Df(x)3Bx2+Df′(x)⇒3x3+1−30x2f(x)[10f(x)−x]=−A3x2−f′(x) where B=1,D=−1.⇒3x3+1−30x2f(x)=−A3x2−f′(x)[10f(x)−x]⇒3x3+f(x)−30x2f(x)=A3x3−xf′(x)−30x2f(x)+10f′(x)f(x)⇒A=1.∴A+B+D=1+1−1=1.

If 5(f(x))2=xf(x)+30 and ∫3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2dx=ABx3+Df(x)+C then A+B+D=

∫3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2dx=ABx3+Df(x)+C⇒3x3+1−30x2f(x)[10f(x)−x]x3−f(x)2=−ABx3+Df(x)3Bx2+Df′(x)⇒3x3+1−30x2f(x)[10f(x)−x]=−A3x2−f′(x) where B=1,D=−1.⇒3x3+1−30x2f(x)=−A3x2−f′(x)[10f(x)−x]⇒3x3+f(x)−30x2f(x)=A3x3−xf′(x)−30x2f(x)+10f′(x)f(x)⇒A=1.∴A+B+D=1+1−1=1.

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If $\begin{array}{l} 5 (f (x))^{2} = x f (x) + 30 and \int \frac{[(3 x^{3} + (1 - 30 x^{2}) f (x)]}{[10 f (x) - x] {[x^{3} - f (x)]}^{2}} d x = \frac{A}{B x^{3} + D f (x)} + C \end{array}$ then $A + B + D =$

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Detailed Solution

$\begin{array}{l} \int \frac{[3 x^{3} + (1 - 30 x^{2}) f (x)]}{[10 f (x) - x] {[x^{3} - f (x)]}^{2}} d x = \frac{A}{B x^{3} + D f (x)} + C \\ \Rightarrow \frac{[3 x^{3} + (1 - 30 x^{2}) f (x)]}{[10 f (x) - x] {[x^{3} - f (x)]}^{2}} = \frac{- A}{B x^{3} + D f (x)} [3 B x^{2} + D f^{'} (x)] \\ \Rightarrow \frac{[3 x^{3} + (1 - 30 x^{2}) f (x)]}{[10 f (x) - x]} = - A [3 x^{2} - f^{'} (x)] where B = 1, D = - 1 . \\ \Rightarrow 3 x^{3} + (1 - 30 x^{2}) f (x) = - A [3 x^{2} - f^{'} (x)] [10 f (x) - x] \\ \Rightarrow 3 x^{3} + f (x) - 30 x^{2} f (x) = A [3 x^{3} - x f^{'} (x) - 30 x^{2} f (x) + 10 f^{'} (x) f (x)] \Rightarrow A = 1 . \\ ∴ A + B + D = 1 + 1 - 1 = 1 . \end{array}$