What is the integral of ∫1√x+1−√x?

1 Answer
Ken C.
Feb 3, 2016

∫1√x+1−√xdx=23(x+1)32+23(x)32+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 √2−1 is √2+1 and the conjugate of √5+√3 is √5−√3. The interesting thing about conjugates is when you multiply them together, you get an interesting result. Look at the product of √2−1 and √2+1:
(√2−1)(√2+1)=2+√2−√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 √x+1−√x. The conjugate of this is √x+1+√x. Let's see what happens when we multiply our fraction by √x+1+√x√x+1+√x (note this is the same thing as multiplying by one):
∫1√x+1−√xdx=∫√x+1+√x√x+1+√x⋅1√x+1−√xdx

=∫√x+1+√x(√x+1+√x)(√x+1−√x)dx

Doing some conjugate multiplication in the denominator,
=∫√x+1+√x(x+1)−(√x)(√x+1)+(√x)(√x+1)−(x)dx

The middle terms cancel out, leaving us with
=∫√x+1+√x(x+1)−(x)dx

And then the xs cancel out, so we have
=∫√x+1+√x1dx

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

Using the properties of integrals,
=∫√x+1dx+∫√xdx

And because √x=x12,
=∫(x+1)12dx+∫(x)12dx

Now all we have is a case of the reverse power rule in both integrals, making them simplify to
∫1√x+1−√xdx=23(x+1)32+23(x)32+C

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

Related questions
See all questions in Integrals of Rational Functions
Impact of this question
2143 views around the world
You can reuse this answer
Creative Commons License