Find $\int\frac{dx}{2+\sqrt{x}}$ (using Integration by Substitution)

Question

I used the substitution:

$u=x$

$du=dx$

$2+\sqrt{u}=2+\sqrt{x}$

I then substituted the u into the equation:

$\int\frac{1}{2+\sqrt{u}}du$

$=\int{(2+\sqrt{u})^{-1}du}$

I'm not too sure how to integrate this?

The solution which my lecturer gave is much different to this. His example is:

$x=u^2$. Then $\frac{dx}{du}=2u$

$\int{\frac{dx}{2+\sqrt{u}}}=\int{\frac{2u}{2+u}du}$

After some thinking, I understand this but I think it would be easier if I knew how to integrate $=\int{(2+\sqrt{u})^{-1}du}$

Thank you very much for your help. I really hope I have been clear enough.

Answer 1

Do this substitution
Let $$u=\frac{1}{2+\sqrt{x}}$$
$$x=\frac{(1-2u)^2}{u^2}$$ Then $$\frac{du}{dx}=\frac{-1}{2\sqrt{x}(2+\sqrt{x})^2}$$
Or $$dx=-2\sqrt{x}(2+\sqrt{x})^2du$$
Substitute these values into the integration.$$\int -2(1-2u)du$$ $$\int (4u-2)du$$
Can you take it from here?

Answer 2

I think your question is why did we cannot integrate diretly $\int{(2+\sqrt{u})^{-1}du}$.

Firstly, notice that your substitution did not alter the integral in any way, you just changed the $x$ into a $u$.

Now when we integrate, we are just finding a function that when differentiated will gives us that same integrand ($\pm$ a constant). This is the Fundamental Theorem of Calculus.

So, can you, from the top of your head think of a function whose derivative is $(2+\sqrt{x})^{-1}$?

It's possible, and calculus students have to know some basic trigonometric, logarithmic or rational integrands. Of course you could try and do it the long way, constructing for example a Riemann Integral.

So we are normally trying to reduce a problem, via some algebra, into one we already know the answer to. So in this case you might not know how to integrate when there is a square root on the denominator but we are trying to manipulate it into something we do know.

Using your teachers method we have:

$$\int\frac{dx}{2+\sqrt{x}}, \text{Let} x=u^2, \int{\frac{2u}{2+u}du} = \int 2 - \frac{4}{2+u} du$$

Now, do we know how to integrate this? Well I hope you do, we have reduced the problem into some common integrands, $\int 2 du= 2u$ and $\int \frac{4}{2+u} du= 4\ln (2+u)$.

So putting everything together and changing back to $x$ we get:

$$\int\frac{dx}{2+\sqrt{x}}= 2\sqrt x - 4\ln(2+\sqrt{x})$$