If x({x^4} + 1)phi (x) = 1, then int_1^2 {phi | vedclass

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(A) Given $x({x^4} + 1)\phi (x) = 1,$ we have $\phi (x) = \frac{1}{{x({x^4} + 1)}}.$ We can write $\phi (x)$ using partial fractions as: $\phi (x) = \

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

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(A) Given $x({x^4} + 1)\phi (x) = 1,$ we have $\phi (x) = \frac{1}{{x({x^4} + 1)}}.$ We can write $\phi (x)$ using partial fractions as: $\phi (x) = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}.$ Now,we integrate $\phi (x)$ from $1$ to $2$: $\int_1^2 {\phi (x)\,dx = \int_1^2 {\left( {\frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}} \right)\,dx} }.$ $= \left[ \log |x| \right]_1^2 - \int_1^2 {\frac{{{x^3}}}{{{x^4} + 1}}\,dx}.$ Let $u = {x^4} + 1,$ then $du = 4{x^3}\,dx,$ so ${x^3}\,dx = \frac{1}{4}du.$ When $x = 1, u = 2.$ When $x = 2, u = 17.$ $= (\log 2 - \log 1) - \frac{1}{4} \int_2^{17} {\frac{1}{u}\,du} = \log 2 - \frac{1}{4} \left[ \log u \right]_2^{17}.$ $= \log 2 - \frac{1}{4} (\log 17 - \log 2) = \log 2 - \frac{1}{4} \log 17 + \frac{1}{4} \log 2.$ $= \frac{5}{4} \log 2 - \frac{1}{4} \log 17 = \frac{1}{4} (5 \log 2 - \log 17) = \frac{1}{4} \log \left( \frac{2^5}{17} \right) = \frac{1}{4} \log \frac{32}{17}.$ Thus,the correct option is $A$.

If $x({x^4} + 1)\phi (x) = 1,$ then $\int_1^2 {\phi (x)\,dx = } $

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