∫limits_1^2dx/(x≤ft(x^4+1)) = ?

\(\int\limits_1^2\frac{dx}{x\left(x^4+1\right)}=\int\limits_1^2\frac{dx}{x^5(1+x^{-4})}\)   

Let $t= 1+x^{-4}$

$ x=1 \rightarrow , t=2 $

$x=2 , \rightarrow t=\frac{17}{16}$

 

$-4x^{-5}dx=dt$

$\int\limits_2^{17/16}\frac{-1}{4t}dt=-\frac{1}{4}[ln\, t]_2^{17/16}⇒-\frac{1}{4}[ln\frac{17}{16}-ln2]$

$=\frac{1}{4}ln[\frac{32}{17}]⇒\frac{1}{4}log[\frac{32}{17}]$

So, option D is correct.