How do you integrate #{x/(sqrt(4+4x^2))} dx# from 0 to 2?

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Answer 1 · 2017-02-01T06:32:16

The answer is #=1/2(sqrt5-1)#

We need
#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Let #u=4+4x^2#

#du=8xdx#,

#xdx=1/8du#

Therefore,

#int(xdx)/sqrt(4+4x^2)=1/8int(du)/sqrtu=1/8u^(-1/2+1)/(1/2)#

#=1/4sqrtu=1/4sqrt(4+4x^2)=1/2sqrt(1+x^2)#

so,

#int_0^2(xdx)/sqrt(4+4x^2)=[1/2sqrt(1+x^2)]_0^2#

#=(sqrt5/2)-(1/2)#

#=1/2(sqrt5-1)#