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

Question

Charles Anctil · Accepted Answer

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#

Evelyn Agard · Answer

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# Power rule: #=((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!

Scarlett Allison · Answer

undefined To integrate $ \frac{1}{\sqrt{x+1} - \sqrt{x}} $, first rationalize the denominator by multiplying the expression by its conjugate:

$ \frac{1}{\sqrt{x+1} - \sqrt{x}} \cdot \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}} $

This simplifies to:

$ \frac{\sqrt{x+1} + \sqrt{x}}{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})} $

Now, you can simplify the denominator:

$ \frac{\sqrt{x+1} + \sqrt{x}}{(x+1) - x} $

$ = \frac{\sqrt{x+1} + \sqrt{x}}{1} $

$ = \sqrt{x+1} + \sqrt{x} $

Now, you can integrate term by term:

$ \int (\sqrt{x+1} + \sqrt{x}) \, dx $

$ = \int \sqrt{x+1} \, dx + \int \sqrt{x} \, dx $

Using substitution and integration rules:

$ = \frac{2}{3}(x+1)^{\frac{3}{2}} + \frac{2}{3}x^{\frac{3}{2}} + C $

Where $ C $ is the constant of integration.