How do you evaluate the integral #int sqrt(e^x+1)#?

Question

28312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-23T06:28:24

Use the substitution #sqrt(e^x+1)=u#.

Let

#I=intsqrt(e^x+1)dx#

Apply the substitution #sqrt(e^x+1)=u#:

#I=2int(u^2)/(u^2-1)du#

Rearrange:

#I=2int(1+1/(u^2-1))du#

Factorize the denominator:

#I=2int(1+1/((u-1)(u+1)))du#

Apply partial fraction decomposition:

#I=int(2+1/(u-1)-1/(u+1))du#

Integrate term by term:

#I=2u+ln|u-1|-ln|u+1|+C#

Reverse the substitution:

#I=2sqrt(e^x+1)+ln|(sqrt(e^x+1)-1)/(sqrt(e^x+1)+1)|+C#