Given that int frac{1}{x^2+a^2} dx = frac{1}{ | vedclass

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(A) We have $I = \int \frac{1}{x^4+3x^2+1} dx$. Dividing numerator and denominator by $x^2$,we get: $I = \int \frac{1/x^2}{x^2 + 1/x^2 + 3} dx$. We ca

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$3$

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(A) We have $I = \int \frac{1}{x^4+3x^2+1} dx$. Dividing numerator and denominator by $x^2$,we get: $I = \int \frac{1/x^2}{x^2 + 1/x^2 + 3} dx$. We can write $1/x^2$ as $\frac{1}{2} \left[ (1 + 1/x^2) - (1 - 1/x^2) ight]$. $I = \frac{1}{2} \int \frac{1 + 1/x^2}{(x - 1/x)^2 + 5} dx - \frac{1}{2} \int \frac{1 - 1/x^2}{(x + 1/x)^2 + 1} dx$. Using the substitution $u = x - 1/x$ $(du = (1 + 1/x^2) dx)$ and $v = x + 1/x$ $(dv = (1 - 1/x^2) dx)$: $I = \frac{1}{2} \int \frac{du}{u^2 + (\sqrt{5})^2} - \frac{1}{2} \int \frac{dv}{v^2 + 1^2}$. $I = \frac{1}{2 \sqrt{5}} 	an^{-1}\left(\frac{x - 1/x}{\sqrt{5}}ight) - \frac{1}{2} 	an^{-1}\left(\frac{x + 1/x}{1}ight) + k$. $I = \frac{1}{2 \sqrt{5}} 	an^{-1}\left(\frac{x^2 - 1}{\sqrt{5}x}ight) - \frac{1}{2} 	an^{-1}\left(\frac{x^2 + 1}{x}ight) + k$. Comparing with the given form,we get $a = \frac{1}{2 \sqrt{5}}$,$b = \frac{1}{\sqrt{5}}$,$c = -\frac{1}{2}$,and $d = 1$. Now,$5(c + d + ab) = 5\left(-\frac{1}{2} + 1 + \frac{1}{2 \sqrt{5}} \cdot \frac{1}{\sqrt{5}}ight) = 5\left(\frac{1}{2} + \frac{1}{10}ight) = 5\left(\frac{6}{10}ight) = 3$.

Given that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. If $\int \frac{1}{x^4+3x^2+1} dx = a \cdot \tan^{-1}\left(\frac{b(x^2-1)}{x}\right) + c \cdot \tan^{-1}\left(\frac{d(x^2+1)}{x}\right) + k$,where $k$ is a constant of integration,then $5(c+d+ab) = $

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