How do you intagrate this function?

Question

(#1+sqrt((x+1)# ) / #1-sqrt((x+1))#

1200 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T14:09:22

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

Let

#I=int(1+sqrt(x+1))/(1-sqrt(x+1))dx#

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

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

Simplify:

#I=int(6-2u-4/u)du#

Integrate directly:

#I=6u-u^2-4ln|u|+C#

Reverse the substitution:

#I=(5+sqrt(x+1))(1-sqrt(x+1))-4ln|1-sqrt(x+1)|+C#

Simplify and rescale #C#:

#I=-x-4sqrt(x+1)-4ln|1-sqrt(x+1)|+C#