Question

o valor da integral

Answer 1

u=x⁴+1   ....du=4x³ dx   ==>dx=du/4x³

∫4*x³*(x⁴+1) dx

∫4*x³*(x⁴+1)⁶ dx

∫4*x³*(u)⁶ du/4x³

∫(u)⁶ du =u⁷/7

Como u = x⁴+1 

u⁷/7   =(x⁴+1 )⁷/7

                  1 

=(x⁴+1 )⁷/7]       =(1+1)⁷/7  -1/7  =(128-1)/7=18,14

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