Evaluate the following integral ∫_1^2dx/x(1+x^4 )

∫_1^2dx/x(1+x^4 )                let x^4=u →4x^3 dx=du→x^3 dx=du/4
x=1 →u=1,x=2→u=16
∴I=∫_1^2dx/x(1+x^4 )   =∫_1^2(x^3 dx)/(x^4 (1+x^4 ) )=1/4 ∫_1^16du/u(1+u)
=   1/4 ∫_1^16[1/u-1/(u+1)]  du=1/4 [ln⁡u-ln⁡(1+u) ]_1^16
=1/4 [ln⁡16-ln⁡17-(ln⁡1-ln⁡2 )]  =1/4 [ln⁡16-ln⁡17+ln⁡2 ]=1/4  ln⁡(32/17)