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

Mar 31, 2018

The answer is $= 0.72$

#### Explanation:

Calculate the indefinite integral first.

Let $2 x = \tan \left(u\right)$, $\implies$, $2 \mathrm{dx} = {\sec}^{2} \left(u\right) \mathrm{du}$

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

Therefore, the integral is

$I = \int \frac{\mathrm{dx}}{\sqrt{1 + 4 {x}^{2}}} = \frac{1}{2} \int \frac{{\sec}^{2} u \mathrm{du}}{\sec} u = \frac{1}{2} \int \sec u \mathrm{du}$

$= \frac{1}{2} \int \frac{\sec u \left(\sec u + \tan u\right) \mathrm{du}}{\sec u + \tan u}$

Let $v = \sec u + \tan u$, $\implies$, $\mathrm{dv} = \left({\sec}^{2} u + \sec u \tan u\right) \mathrm{du}$

Therefore,

$I = \frac{1}{2} \int \frac{\mathrm{dv}}{v} = \ln v$

$= \frac{1}{2} \ln \left(\sec u + \tan u\right)$

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

Then, the definite integral is

${\int}_{0}^{1} \frac{\mathrm{dx}}{\sqrt{1 + 4 {x}^{2}}} = {\left[\frac{1}{2} \ln \left(| \sqrt{1 + 4 {x}^{2}} + 2 x |\right)\right]}_{0}^{1}$

$= \left(\frac{1}{2} \ln \left(2 + \sqrt{5}\right)\right) - \left(\frac{1}{2} \ln \left(1\right)\right)$

$= 0.72$

Mar 31, 2018

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

#### Explanation:

${\int}_{0}^{1} \frac{1}{\sqrt{1 + 4 {x}^{2}}} . \mathrm{dx}$
${\left[{1}^{-} \left(\frac{1}{2}\right)\right]}_{0}^{1} + {\left[{\left(4 {x}^{2}\right)}^{- \frac{1}{2}}\right]}_{0}^{1}$
$1 + {\left[4 {\left(1\right)}^{2} - 4 {\left(0\right)}^{2}\right]}^{-} \left(\frac{1}{2}\right)$
$1 + {\left(4\right)}^{- \frac{1}{2}}$
$1 + \frac{1}{\sqrt{4}}$
$1 + \frac{1}{2}$
=)$\frac{3}{2}$