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

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(D) આપણી પાસે $I = \int \frac{1}{x^4+8x^2+9} dx$ છે. અંશ અને છેદને $x^2$ વડે ભાગતા,આપણને મળે $I = \frac{1}{6} \int \frac{(1+3/x^2)}{(x-3/x)^2+14} dx -

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$\sqrt{2}+1$

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(D) આપણી પાસે $I = \int \frac{1}{x^4+8x^2+9} dx$ છે. અંશ અને છેદને $x^2$ વડે ભાગતા,આપણને મળે $I = \frac{1}{6} \int \frac{(1+3/x^2)}{(x-3/x)^2+14} dx - \frac{1}{6} \int \frac{(1-3/x^2)}{(x+3/x)^2+2} dx$. ધારો કે $t = x-3/x$ અને $u = x+3/x$. તેથી $dt = (1+3/x^2) dx$ અને $du = (1-3/x^2) dx$. $I = \frac{1}{6} \int \frac{dt}{t^2+(\sqrt{14})^2} - \frac{1}{6} \int \frac{du}{u^2+(\sqrt{2})^2}$. $I = \frac{1}{6} \left[ \frac{1}{\sqrt{14}} 	an^{-1} \left( \frac{x-3/x}{\sqrt{14}} ight) - \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{x+3/x}{\sqrt{2}} ight) ight] + c$. આપેલ સ્વરૂપ સાથે સરખાવતા,$k=6$,$f(x) = \frac{x-3/x}{\sqrt{14}}$,અને $g(x) = \frac{x+3/x}{\sqrt{2}}$. હવે,$f(\sqrt{3}) = \frac{\sqrt{3}-3/\sqrt{3}}{\sqrt{14}} = 0$ અને $g(1) = \frac{1+3/1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$. તેથી,$\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} = \sqrt{\frac{6}{2} + 0 + 2\sqrt{2}} = \sqrt{3+2\sqrt{2}} = \sqrt{(\sqrt{2}+1)^2} = \sqrt{2}+1$.

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

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