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 = $3/2$

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

how do you integrate int_0^1 1/sqrt(1+4x^2) dxint_0^1 1/sqrt(1+4x^2) dx ?