how do you integrate #int_0^1 1/sqrt(1+4x^2) dx# ?

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Answer 1 · 2018-03-31T08:28:25

The answer is #=0.72#

Calculate the indefinite integral first.

Let #2x=tan(u)#, #=>#, #2dx=sec^2(u)du#

#sqrt(1+4x^2)=sqrt(1+tan^2(u))=secu#

Therefore, the integral is

#I=int(dx)/sqrt(1+4x^2)=1/2int(sec^2udu)/secu=1/2intsecudu#

#=1/2int(secu(secu+tanu)du)/(secu+tanu)#

Let #v=secu+tanu#, #=>#, #dv=(sec^2u+secutanu)du#

Therefore,

#I=1/2int(dv)/(v)=lnv#

#=1/2ln(secu+tanu)#

#=1/2ln(|sqrt(1+4x^2)+2x|)+C#

Then, the definite integral is

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

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

#=0.72#

Answer 2 · 2018-03-31T08:30:05

answer = #3/2#

#int_0^1 1/sqrt(1+4x^2).dx#
#[1^-(1/2)]_0^1 + [(4x^2)^(-1/2)]_0^1#
#1+[4(1)^2-4(0)^2]^-(1/2)#
#1+(4)^(-1/2)#
#1+1/sqrt4#
#1+1/2#
=)#3/2#