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

1 Answer
Ken C.
Feb 3, 2016

#int1/(sqrt(x+1)-sqrt(x))dx=2/3(x+1)^(3/2)+2/3(x)^(3/2)+C#

Explanation:

When I first looked at this one I thought it would be quite difficult, but in fact all we have to do is remember a basic technique from algebra: conjugate multiplication.

You may recall that the conjugate of a number is what you get when you switch the sign in the middle of a binomial. For example, the conjugate of #sqrt(2)-1# is #sqrt(2)+1# and the conjugate of #sqrt(5)+sqrt(3)# is #sqrt(5)-sqrt(3)#. The interesting thing about conjugates is when you multiply them together, you get an interesting result. Look at the product of #sqrt(2)-1# and #sqrt(2)+1#:
#(sqrt(2)-1)(sqrt(2)+1)=2+sqrt(2)-sqrt(2)-1=2-1=1#.

We can see that even though our original expression contained square roots, our answer didn't - which, as we will see, is very useful.

In the denominator of our integral, we have #sqrt(x+1)-sqrt(x)#. The conjugate of this is #sqrt(x+1)+sqrt(x)#. Let's see what happens when we multiply our fraction by #(sqrt(x+1)+sqrt(x))/(sqrt(x+1)+sqrt(x))# (note this is the same thing as multiplying by one):
#int1/(sqrt(x+1)-sqrt(x))dx=int(sqrt(x+1)+sqrt(x))/(sqrt(x+1)+sqrt(x))*1/(sqrt(x+1)-sqrt(x))dx#

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

Doing some conjugate multiplication in the denominator,
#=int(sqrt(x+1)+sqrt(x))/((x+1)-(sqrt(x))(sqrt(x+1))+(sqrt(x))(sqrt(x+1))-(x))dx#

The middle terms cancel out, leaving us with
#=int(sqrt(x+1)+sqrt(x))/((x+1)-(x))dx#

And then the #x#s cancel out, so we have
#=int(sqrt(x+1)+sqrt(x))/(1)dx#

We see the beauty of conjugate multiplication now, as our seemingly complicated integral is now reduced to
#=int(sqrt(x+1)+sqrt(x))dx#

Using the properties of integrals,
#=intsqrt(x+1)dx+intsqrt(x)dx#

And because #sqrt(x)=x^(1/2)#,
#=int(x+1)^(1/2)dx+int(x)^(1/2)dx#

Now all we have is a case of the reverse power rule in both integrals, making them simplify to
#int1/(sqrt(x+1)-sqrt(x))dx=2/3(x+1)^(3/2)+2/3(x)^(3/2)+C#

And that, ladies and gentlemen, is our final answer.

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