Question

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

`int_ln2^ln3e^x/sqrt(e^(2x)+1)dx`

Answer 1

`int_ln2^ln3e^x/sqrt(e^(2x)+1)dx=ln((sqrt10+3)/(sqrt5+2))`

Explanation:

For `int_ln2^ln3e^x/sqrt(e^(2x)+1)dx`

let `t=e^x`, then `dt=e^xdx`

and `int_2^3 1/sqrt(t^2+1)dt`

= `[ln|sqrt(t^2+1)+t|]_2^3`

= `ln|sqrt10+3|-ln|sqrt5+2|`

= `ln((sqrt10+3)/(sqrt5+2))`

Answer 2

The answer is `=0.37`

Explanation:

Calculate the indefinite integral first

Perform the substitution

`u=e^x`, `=>`, `du=e^xdx`

Therefore,

`I=int(e^xdx)/(sqrt(e^(2x)+1))`

`=int(du)/(sqrt(u^2+1))`

Let `u=tantheta`, `=>`, `du=sec^2thetad theta`

`sqrt(u^2+1)=sqrt(tan^2theta+1)=sectheta`

Therefore,

`I=int(sec^2thetad theta)/(sectheta)=int sec thetad theta`

`=int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)`

Let,

`v=sectheta+tantheta`, `=>`,

`dv=(secthetatantheta+sec^2theta)d theta`

So,

`I=int(dv)/(v)`

`=ln(v)`

`=ln(sectheta+tantheta)`

`=ln(sqrt(1+u^2)+u)`

`=ln(sqrt(1+e^(2x))+e^x)+C`

Now, compute the definite integral

`int_ln2^ln3(e^xdx)/(sqrt(e^(2x)+1))= [ln(sqrt(1+e^(2x))+e^x)]_ln2^ln3`

`=(ln(sqrt(1+e^(2ln3))+e^ln3))-(ln(sqrt(1+e^(2ln2))+e^ln2))`

`=(ln(sqrt10+3))-(ln(sqrt5+2))`

`=0.37`

Answer 3

`I=ln|(3+sqrt10)/(2+sqrt5)|`

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_ln2^ln3(e^x/sqrt(e^(2x)+1))dx`

Let, `e^x=t=>e^xdx=dt,and`

`x=ln2=>t=e^ln2=2,and x=ln3=>t=e^ln3=3`

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

`=>I=[ln|t+sqrt(t^2+1)|]_2^3`

`=>I= ln|3+sqrt(3^2+1)|-ln|2+sqrt(2^2+1)|`

`=>I=ln|3+sqrt10|-ln|2+sqrt5|`

`=>I=ln|(3+sqrt10)/(2+sqrt5)|`