How do you integrate int_ln2^ln3(e^x/sqrt(e^(2x)+1)dx ?

${\int}_{\ln} {2}^{\ln} 3 {e}^{x} / \sqrt{{e}^{2 x} + 1} \mathrm{dx}$

Mar 23, 2018

${\int}_{\ln} {2}^{\ln} 3 {e}^{x} / \sqrt{{e}^{2 x} + 1} \mathrm{dx} = \ln \left(\frac{\sqrt{10} + 3}{\sqrt{5} + 2}\right)$

Explanation:

For ${\int}_{\ln} {2}^{\ln} 3 {e}^{x} / \sqrt{{e}^{2 x} + 1} \mathrm{dx}$

let $t = {e}^{x}$, then $\mathrm{dt} = {e}^{x} \mathrm{dx}$

and ${\int}_{2}^{3} \frac{1}{\sqrt{{t}^{2} + 1}} \mathrm{dt}$

= ${\left[\ln | \sqrt{{t}^{2} + 1} + t |\right]}_{2}^{3}$

= $\ln | \sqrt{10} + 3 | - \ln | \sqrt{5} + 2 |$

= $\ln \left(\frac{\sqrt{10} + 3}{\sqrt{5} + 2}\right)$

Mar 23, 2018

The answer is $= 0.37$

Explanation:

Calculate the indefinite integral first

Perform the substitution

$u = {e}^{x}$, $\implies$, $\mathrm{du} = {e}^{x} \mathrm{dx}$

Therefore,

$I = \int \frac{{e}^{x} \mathrm{dx}}{\sqrt{{e}^{2 x} + 1}}$

$= \int \frac{\mathrm{du}}{\sqrt{{u}^{2} + 1}}$

Let $u = \tan \theta$, $\implies$, $\mathrm{du} = {\sec}^{2} \theta d \theta$

$\sqrt{{u}^{2} + 1} = \sqrt{{\tan}^{2} \theta + 1} = \sec \theta$

Therefore,

$I = \int \frac{{\sec}^{2} \theta d \theta}{\sec \theta} = \int \sec \theta d \theta$

$= \int \frac{\sec \theta \left(\sec \theta + \tan \theta\right) d \theta}{\sec \theta + \tan \theta}$

Let,

$v = \sec \theta + \tan \theta$, $\implies$,

$\mathrm{dv} = \left(\sec \theta \tan \theta + {\sec}^{2} \theta\right) d \theta$

So,

$I = \int \frac{\mathrm{dv}}{v}$

$= \ln \left(v\right)$

$= \ln \left(\sec \theta + \tan \theta\right)$

$= \ln \left(\sqrt{1 + {u}^{2}} + u\right)$

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

Now, compute the definite integral

${\int}_{\ln} {2}^{\ln} 3 \frac{{e}^{x} \mathrm{dx}}{\sqrt{{e}^{2 x} + 1}} = {\left[\ln \left(\sqrt{1 + {e}^{2 x}} + {e}^{x}\right)\right]}_{\ln} {2}^{\ln} 3$

$= \left(\ln \left(\sqrt{1 + {e}^{2 \ln 3}} + {e}^{\ln} 3\right)\right) - \left(\ln \left(\sqrt{1 + {e}^{2 \ln 2}} + {e}^{\ln} 2\right)\right)$

$= \left(\ln \left(\sqrt{10} + 3\right)\right) - \left(\ln \left(\sqrt{5} + 2\right)\right)$

$= 0.37$

Mar 23, 2018

$I = \ln | \frac{3 + \sqrt{10}}{2 + \sqrt{5}} |$

Explanation:

We know that,

color(red)(e^(log_eX)=X=>e^lnX=X

color(red)(int1/(sqrt(X^2+k))dx=ln|X+sqrt(X^2+k)|+c

Here,

$I = {\int}_{\ln} {2}^{\ln} 3 \left({e}^{x} / \sqrt{{e}^{2 x} + 1}\right) \mathrm{dx}$

Let, ${e}^{x} = t \implies {e}^{x} \mathrm{dx} = \mathrm{dt} , \mathmr{and}$

$x = \ln 2 \implies t = {e}^{\ln} 2 = 2 , \mathmr{and} x = \ln 3 \implies t = {e}^{\ln} 3 = 3$

:.I=int_2^3(dt)/sqrt(t^2+1

$\implies I = {\left[\ln | t + \sqrt{{t}^{2} + 1} |\right]}_{2}^{3}$

$\implies I = \ln | 3 + \sqrt{{3}^{2} + 1} | - \ln | 2 + \sqrt{{2}^{2} + 1} |$

$\implies I = \ln | 3 + \sqrt{10} | - \ln | 2 + \sqrt{5} |$

$\implies I = \ln | \frac{3 + \sqrt{10}}{2 + \sqrt{5}} |$