# 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