Integral of a function

1. Feb 26, 2006

gulsen

$$\int \frac{dx}{\sqrt{1/x + C}}$$ where C is a constant. Any ideas?

2. Feb 26, 2006

arildno

$$u=\sqrt{\frac{x}{1+Cx}}$$
Try and integrate with respect to u; that usually works in cases like this.

3. Feb 27, 2006

HallsofIvy

Is that $\frac{1}{x}+ C$ or $\frac{1}{x+C}$?

4. Feb 27, 2006

benorin

The former, viz. $$\int \frac{1}{\sqrt{\frac{1}{x}+C}} dx = \int\sqrt{\frac{x}{1+Cx}} dx,$$ is nasty looking and the later is quite simple, of chourse it is just $$\int\sqrt{x+C}dx = \frac{2}{3}(x+C)^{\frac{3}{2}}+C_1$$

5. Feb 27, 2006

robert spicuzza

Maybe a little longer approach,
But this is the way I’d approach this in a cal 2 class

First rewrite integral sqrt(X)/sqrt(1+CX) dx

Now let 1/x =u^2
dx=-2/(1/u^3)du
also let a^2 = c to make notation a little easier to work with.

This will yield an integral that look likes within constants
1/(u^3 sqrt(u^2+a^2) ) du

Now use a standard trig substitution let tan(angle)= a^2 u^2
the top numerator is just sec^2(angle) *d (angle)
the bottom portion is tan(angle)^3 * sec(angle)

Now simplify
We are integrating to within a constants sec(angle)/tan^2(angle) * d(angle)

Still have some simplification to do using the definition of tan = sin/cos
And sec=1/cos.

The form reduces to cos(angle) d(angle) / sin^2 (angle)
The integral is of this form is 1/sin^3(angle) and now all you need to do is all the back substitution.

And you can now begine to see the (2/3)(X+C)^(3/2) form reported by Benorin

If I was teaching cal 2 ( I’m doing cal 1) this semester I’d consider this a great take home problem for credit.

A rewrite would be nice if someone has a program which writes in a better math notation.

Again this is a great problem!

6. Feb 27, 2006

arildno

$$u=\sqrt{\frac{x}{1+Cx}}\to{u}^{2}=\frac{x}{1+Cx}\to{x}=\frac{u^{2}}{1-Cu^{2}}$$

We therefore have:
$$\frac{dx}{du}=\frac{2u(1-Cu^{2})+2Cu^{3}}{(1-Cu^{2})^{2}}=\frac{2u}{(1-Cu^{2})^{2}}$$
by which it follows:
$$\int\sqrt{\frac{x}{1+Cx}}dx=\int\frac{2u^{2}}{(1-Cu^{2})^{2}}du$$
The latter integral is easily calculated.

Last edited: Feb 27, 2006
7. Feb 27, 2006

gulsen

Easily??
Well, the previous ones seem to be wrong, I got this monster from mathematica:

$$\frac{cx+1}{c \sqrt{c + \frac{1}{x}}} - \frac{\sqrt{cx+1} asinh {\sqrt{cx}} }{c^{3/2} \sqrt{c + \frac{1}{x}} \sqrt {x}}$$

BTW, how did you guys derieved $$\int\sqrt{x+C}dx$$ from $$\int\sqrt{\frac{x}{1+Cx}} dx$$?

Last edited: Feb 27, 2006
8. Mar 1, 2006

VietDao29

No, we did not say that:
$$\int \sqrt{x + C} dx = \int \sqrt{\frac{x}{1 + Cx}} dx$$
What benorin says is just that, if the original integral is:
$$\int \frac{dx}{\sqrt{\frac{1}{x + C}}}$$, then it can be rearranged to give:
$$\int \frac{dx}{\sqrt{\frac{1}{x + C}}} = \int \sqrt{x + C} dx$$.
:)

9. Mar 1, 2006

gulsen

Whoops! :)
Hmm, well then, what if it was $$\int\sqrt{\frac{x}{1+Cx}} dx$$? If what mathematica says is true, then it's gonna be a BIG headache to solve. But I want to see what this headache is!

10. Mar 1, 2006

HallsofIvy

Partial fractions is not a big headache. Tedious perhaps.

11. Mar 2, 2006

arildno

Hint:
Use integration by parts first:
$$\int\frac{2u^{2}}{(1-Cu^{2})^{2}}du=\frac{u}{C(1-Cu^{2})}-\frac{1}{C}\int\frac{du}{1-Cu^{2}}$$
Use partial fractions on the latter integral, and you're done

Last edited: Mar 2, 2006
12. Mar 8, 2006

misskitty

How can you write 1 + Cx when the original is 1/x + C? Pardon my ingnorance on this subject, but we just started these last week.

13. Mar 8, 2006

arildno

Very good question, misskitty!
Remember that the original integrand is $\frac{1}{\sqrt{\frac{1}{x}+C}}$
Now, we may rewrite this as:
$$\frac{1}{\sqrt{\frac{1}{x}+C}}=\sqrt{\frac{1}{\frac{1}{x}+C}}=\sqrt{\frac{x}{x}*\frac{1}{\frac{1}{x}+C}}=\sqrt{\frac{x}{Cx+1}}$$

14. Mar 8, 2006

misskitty

OH!!! So you CAN do that because all you've done is multiplied by a name for one inside the radical! That would have made one of my problems so much easier. Thanks for explaining.

~Kitty

15. Mar 8, 2006

misskitty

Also what's the purpose of changing the variable from x to u? What's the difference between integrating with respect to x or respect to u? Its still integrating with respect to a variable.

~Kitty

16. Mar 8, 2006

arildno

Note that the u-substitution transforms your integral from an ugly expression involving a square root into a rational function (i.e, whose numerator and denominator are polynomials).
Since we know how to integrate rational functions, we've just made our original problem easily solvable..

17. Mar 8, 2006

misskitty

So by replacing it with u then you can just deal with the numbers without the square root and then once everything is worked out then you can take the square root. Do I have this correct?

~Kitty

18. Mar 8, 2006

arildno

Yes, having found your anti-derivative in u, just replace every u through $u=\sqrt{\frac{x}{Cx+1}}$, and you're done.

19. Mar 8, 2006

misskitty

I get it now! Yay. Thanks for the patience to explain it.

~Kitty