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

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

3 Answers
Mar 23, 2018

Answer:

#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))#

Mar 23, 2018

Answer:

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#

Mar 23, 2018

Answer:

#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)|#