How do you integrate #int 1 / (sqrt(x+1) - sqrt(x)) #?

Question

2657 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-02T20:27:00

The integral is equal to #2/3((x+1)^(3/2)+x^(3/2))+C#.

First, rationalize the denominator:

#color(white)=int 1/(sqrt(x+1)-sqrtx)dx#

#=int 1/(sqrt(x+1)-sqrtx)color(red)(*(sqrt(x+1)+sqrtx)/(sqrt(x+1)+sqrtx))dx#

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

#=int (sqrt(x+1)+sqrtx)/((sqrt(x+1))^2-(sqrtx)^2)dx#

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

#=int (sqrt(x+1)+sqrtx)/(color(red)cancelcolor(black)x+1color(red)cancelcolor(black)(color(black)-x))dx#

#=int (sqrt(x+1)+sqrtx)/1dx#

#=int (sqrt(x+1)+sqrtx) dx#

#=intsqrt(x+1)# #dx+intsqrtx# #dx#

#=int(x+1)^(1/2)# #dx+intx^(1/2)# #dx#

#=((x+1)^(1/2+1))/(1/2+1)+(x^(1/2+1))/(1/2+1)#

#=((x+1)^(3/2))/(3/2)+(x^(3/2))/(3/2)#

#=2/3(x+1)^(3/2)+2/3x^(3/2)#

You can factor out the #2/3#, and don't forget to add #C#:

#=2/3((x+1)^(3/2)+x^(3/2))+C#

That's the whole integral. Hope this helped!

Answer 2 · 2018-04-02T20:29:45

See below.

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

hence

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