If int frac{1}{x^4+8 x^2+9} d x = frac{1}{k} | vedclass

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(D) We have $I = \int \frac{1}{x^4+8x^2+9} dx$. Dividing numerator and denominator by $x^2$, we get $I = \int \frac{1+3/x^2}{x^2+8+9/x^2} dx - \int \f

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$\sqrt{2}+1$

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(D) We have $I = \int \frac{1}{x^4+8x^2+9} dx$. Dividing numerator and denominator by $x^2$, we get $I = \int \frac{1+3/x^2}{x^2+8+9/x^2} dx - \int \frac{3/x^2-1}{x^2+8+9/x^2} dx$ is not direct, so we use: $I = \frac{1}{6} \int \frac{(1+3/x^2)}{(x-3/x)^2+14} dx - \frac{1}{6} \int \frac{(1-3/x^2)}{(x+3/x)^2+2} dx$. Let $t = x-3/x$ and $u = x+3/x$. Then $dt = (1+3/x^2) dx$ and $du = (1-3/x^2) dx$. $I = \frac{1}{6} \int \frac{dt}{t^2+(\sqrt{14})^2} - \frac{1}{6} \int \frac{du}{u^2+(\sqrt{2})^2}$. $I = \frac{1}{6} \left[ \frac{1}{\sqrt{14}} 	an^{-1} \left( \frac{x-3/x}{\sqrt{14}} ight) - \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{x+3/x}{\sqrt{2}} ight) ight] + c$. Comparing with the given form, $k=6$, $f(x) = \frac{x-3/x}{\sqrt{14}}$, and $g(x) = \frac{x+3/x}{\sqrt{2}}$. Now, $f(\sqrt{3}) = \frac{\sqrt{3}-3/\sqrt{3}}{\sqrt{14}} = 0$ and $g(1) = \frac{1+3/1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$. Thus, $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} = \sqrt{\frac{6}{2} + 0 + 2\sqrt{2}} = \sqrt{3+2\sqrt{2}} = \sqrt{(\sqrt{2}+1)^2} = \sqrt{2}+1$.

If $\int \frac{1}{x^4+8 x^2+9} d x = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$, then $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$

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