How do you integrate #int xroot3(4+x^2)# from [0,2]?

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Answer 1 · 2016-11-11T15:24:53

The answer is #=6-3/2^(1/3)#

Use the substitution, #u=4+x^2#
so, #du=2xdx#
#:. intx(4+x^2)^(1/3)dx=intu^(1/3)(du)/2#
#=1/2u^(4/3)/(4/3)=3/8(4+x^2)^(4/3)+C#

So # int_0^2x(4+x^2)^(1/3)dx=[3/8(4+x^2)^(4/3)] _0^2#

#=3/8*(4+4)^(4/3)-3/8*4^(4/3)#
#=3/8*16-3/8*4*4^(1/3)#
#=6-3/2^(1/3)#

Answer 2 · 2016-11-11T15:57:10

Nov 11, 2016

See answer below:
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