If f(x) is a polynomial satisfying f(x).f1x=f(x)+f1x, and f(3) = 82, then ∫f(x)x2+1 dx=

f(x).f1x=f(x)+f1x ⇒f(x)=1±xn f(3)=82=1+81=1+34 ∴f(x)=1+x4 ∫f(x)x2+1=∫x4+1x2+1dx =∫x2-1+2x2+1dx=x33-x+2 tan-1x

If f(x) is a polynomial satisfying f(x).f1x=f(x)+f1x, and f(3) = 82, then ∫f(x)x2+1 dx=

f(x).f1x=f(x)+f1x ⇒f(x)=1±xn f(3)=82=1+81=1+34 ∴f(x)=1+x4 ∫f(x)x2+1=∫x4+1x2+1dx =∫x2-1+2x2+1dx=x33-x+2 tan-1x

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If f(x) is a polynomial satisfying $f (x) . f (\frac{1}{x}) = f (x) + f (\frac{1}{x}),$ and f(3) = 82, then $\int \frac{f (x)}{x^{2} + 1} d x =$

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$x^{3} - x + 2 \tan^{- 1} x + c$

$\frac{x^{3}}{3} - x + 2 \tan^{- 1} x + c$

$\frac{1}{3} x^{3} - x + \tan^{- 1} x + c$

$\frac{1}{3} x^{3} + x + 2 \tan^{- 1} x + c$

answer is C.

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Detailed Solution

$f (x) . f (\frac{1}{x}) = f (x) + f (\frac{1}{x}) \Rightarrow f (x) = 1 \pm x^{n} f (3) = 82 = 1 + 81 = 1 + 3^{4} ∴ f (x) = 1 + x^{4} \int \frac{f (x)}{x^{2} + 1} = \int \frac{x^{4} + 1}{x^{2} + 1} dx = \int (x^{2} - 1 + \frac{2}{x^{2} + 1}) dx = \frac{x^{3}}{3} - x + 2 \tan^{- 1} x$