What is the indefinite integral of #1/(1+sqrt(x+1))#?

1 Answer
May 23, 2018

#int1/(1+sqrt(x+1))dx=2sqrt(x+1)-2ln|1+sqrt(x+1)|+C#

Explanation:

Let

#I=int1/(1+sqrt(x+1))dx#

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

#I=2intu/(1+u)du#

Rearrange:

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

Integrate directly:

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

Reverse the substitution:

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