how do you integrate $\int_{0}^{1} \frac{1}{\sqrt{1 + 4 x^{2}}} d x$ $int_0^1 1/sqrt(1+4x^2) dx$ ?

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how do you integrate $\int_{0}^{1} \frac{1}{\sqrt{1 + 4 x^{2}}} d x$ $int_0^1 1/sqrt(1+4x^2) dx$ ?

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Answer 1 · 2018-03-31T08:28:25

The answer is $= 0.72$ $=0.72$

Explanation:

Calculate the indefinite integral first.

Let $2 x = tan (u)$ $2x=tan(u)$ , $\Rightarrow$ $=>$ , $2 d x = {sec}^{2} (u) d u$ $2dx=sec^2(u)du$

$\sqrt{1 + 4 x^{2}} = \sqrt{1 + {tan}^{2} (u)} = sec u$ $sqrt(1+4x^2)=sqrt(1+tan^2(u))=secu$

Therefore, the integral is

$I = \int \frac{d x}{\sqrt{1 + 4 x^{2}}} = \frac{1}{2} \int \frac{{sec}^{2} u d u}{sec u} = \frac{1}{2} \int sec u d u$ $I=int(dx)/sqrt(1+4x^2)=1/2int(sec^2udu)/secu=1/2intsecudu$

$= \frac{1}{2} \int \frac{sec u (sec u + tan u) d u}{sec u + tan u}$ $=1/2int(secu(secu+tanu)du)/(secu+tanu)$

Let $v = sec u + tan u$ $v=secu+tanu$ , $\Rightarrow$ $=>$ , $d v = ({sec}^{2} u + sec u tan u) d u$ $dv=(sec^2u+secutanu)du$

Therefore,

$I = \frac{1}{2} \int \frac{d v}{v} = ln v$ $I=1/2int(dv)/(v)=lnv$

$= \frac{1}{2} ln (sec u + tan u)$ $=1/2ln(secu+tanu)$

$= \frac{1}{2} ln (∣ ∣ \sqrt{1 + 4 x^{2}} + 2 x ∣ ∣) + C$ $=1/2ln(|sqrt(1+4x^2)+2x|)+C$

Then, the definite integral is

$\int_{0}^{1} \frac{d x}{\sqrt{1 + 4 x^{2}}} = {[\frac{1}{2} ln (∣ ∣ \sqrt{1 + 4 x^{2}} + 2 x ∣ ∣)]}_{0}^{1}$ $int_0^1(dx)/sqrt(1+4x^2)=[1/2ln(|sqrt(1+4x^2)+2x|)]_0^1$

$= (\frac{1}{2} ln (2 + \sqrt{5})) - (\frac{1}{2} ln (1))$ $=(1/2ln(2+sqrt5))-(1/2ln(1))$

$= 0.72$ $=0.72$

Answer 2 · 2018-03-31T08:30:05

answer = $\frac{3}{2}$ $3/2$

Explanation:

$\int_{0}^{1} \frac{1}{\sqrt{1 + 4 x^{2}}} . d x$ $int_0^1 1/sqrt(1+4x^2).dx$
${[1^{- (\frac{1}{2})}]}_{0}^{- (\frac{1}{2})} + {[{(4 x^{2})}^{- \frac{1}{2}}]}_{0}^{- \frac{1}{2}}$ $[1^-(1/2)]_0^1 + [(4x^2)^(-1/2)]_0^1$
$1 + {[4 {(1)}^{2} - 4 {(0)}^{2}]}^{- (\frac{1}{2})}$ $1+[4(1)^2-4(0)^2]^-(1/2)$
$1 + {(4)}^{- \frac{1}{2}}$ $1+(4)^(-1/2)$
$1 + \frac{1}{\sqrt{4}}$ $1+1/sqrt4$
$1 + \frac{1}{2}$ $1+1/2$
=) $\frac{3}{2}$ $3/2$

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how do you integrate $\int_{0}^{1} \frac{1}{\sqrt{1 + 4 x^{2}}} d x$ $int_0^1 1/sqrt(1+4x^2) dx$ ?

2 Answers

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how do you integrate $\int_{0}^{1} \frac{1}{\sqrt{1 + 4 x^{2}}} d x$ $int_0^1 1/sqrt(1+4x^2) dx$ ?