int frac{dx}{x[(log x)^2 + 4log x - 1]} = | vedclass

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(A) Let $\log x = t$. Then,$\frac{1}{x} dx = dt$. Substituting this into the integral,we get: $\int \frac{dt}{t^2 + 4t - 1}$. Completing the square in

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$\frac{1}{2\sqrt{5}}\log \left[ \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right] + c$

Vedclass · Answer

(A) Let $\log x = t$. Then,$\frac{1}{x} dx = dt$. Substituting this into the integral,we get: $\int \frac{dt}{t^2 + 4t - 1}$. Completing the square in the denominator: $t^2 + 4t - 1 = (t^2 + 4t + 4) - 4 - 1 = (t + 2)^2 - 5 = (t + 2)^2 - (\sqrt{5})^2$. Using the standard integral formula $\int \frac{du}{u^2 - a^2} = \frac{1}{2a} \log \left| \frac{u - a}{u + a} \right| + c$: $\int \frac{dt}{(t + 2)^2 - (\sqrt{5})^2} = \frac{1}{2\sqrt{5}} \log \left| \frac{t + 2 - \sqrt{5}}{t + 2 + \sqrt{5}} \right| + c$. Substituting $t = \log x$ back into the expression: $\frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + c$.

$\int \frac{dx}{x[(\log x)^2 + 4\log x - 1]} = $

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