How do you evaluate the integral #int sqrt(e^x+1)#?
1 Answer
May 23, 2018
Use the substitution
Explanation:
Let
#I=intsqrt(e^x+1)dx#
Apply the substitution
#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#