The value of int_1^2 frac{dx}{x(1 + x^4)} is | vedclass

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(D) Let $I = \int_1^2 \frac{dx}{x(1 + x^4)}$.Multiply the numerator and denominator by $x^3$:$I = \int_1^2 \frac{x^3 dx}{x^4(1 + x^4)}$.Let $x^4 = t$,

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$\frac{1}{4}\log \frac{32}{17}$

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(D) Let $I = \int_1^2 \frac{dx}{x(1 + x^4)}$.Multiply the numerator and denominator by $x^3$:$I = \int_1^2 \frac{x^3 dx}{x^4(1 + x^4)}$.Let $x^4 = t$,then $4x^3 dx = dt$,which implies $x^3 dx = \frac{dt}{4}$.When $x = 1$,$t = 1^4 = 1$. When $x = 2$,$t = 2^4 = 16$.Substituting these into the integral:$I = \frac{1}{4} \int_1^{16} \frac{dt}{t(1 + t)}$.Using partial fractions,$\frac{1}{t(1 + t)} = \frac{1}{t} - \frac{1}{1 + t}$.$I = \frac{1}{4} \int_1^{16} \left( \frac{1}{t} - \frac{1}{1 + t} ight) dt$.$I = \frac{1}{4} [\log|t| - \log|1 + t|]_1^{16} = \frac{1}{4} [\log|\frac{t}{1 + t}|]_1^{16}$.$I = \frac{1}{4} (\log \frac{16}{17} - \log \frac{1}{2}) = \frac{1}{4} \log (\frac{16}{17} 	imes 2) = \frac{1}{4} \log \frac{32}{17}$.

The value of $\int_1^2 \frac{dx}{x(1 + x^4)}$ is

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