# Intergration of to functions:

1. Apr 27, 2007

### hhkeep21

Hello, I'm having difficulty intergrating the following two functions:

1) (6x-x^2)^(-1/2)

I'm fairly sure that you need to intergrate by parts for this function, but I can't figure out how to do it

2) e/(e^(2x)-1)

This one I thought could do with inverse trig functions but I'm drawing a blank there also

2. Apr 28, 2007

### Gib Z

1) $$\int \frac{1}{\sqrt{6x-x^2}} dx$$

Whenever I see something like that, I take out as much of the x's from the square root as possible. So lets take out an x^2 :)

$$\int \frac{1}{\sqrt{6x-x^2}} dx = \int \frac{1}{\sqrt{x^2}\sqrt{\frac{6}{x} -1}}$$

Now thats starting to look like something a trig substitution can help :)

I seem to remember $\sec^2 u -1 = \tan^2 u$ :)

So to make it very convenient, the 6/x would be nice to be sec^2 u.

So let x= 6 cos^2 u. Thats a nice substitution :)

2)$$\int \frac{e^x}{(e^x)^2-1} dx$$

u=e^x, du = e^x dx

So it becomes $$\int \frac{1}{u^2-1} du$$

To finish that off, realise the bottom factors by difference of two squares into (u+1)(u-1) and use partial fractions.

Last edited: Apr 28, 2007
3. Apr 28, 2007

### hhkeep21

Thank you so much for your help, I really appreciate it

4. Apr 28, 2007

### VietDao29

A shorter version is:

$$\int \frac{dx}{\sqrt{6x - x ^ 2}} = \int \frac{dx}{\sqrt{- \left( -6x + x ^ 2 \right)}} = \int \frac{dx}{\sqrt{9 - \left( x ^ 2 - 6x + 9 \right)}} = \int \frac{dx}{\sqrt{9 - \left( x - 3 \right) ^ 2}} = ...$$

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Usually, when you encounter the form:
$$\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}}$$
We divide it into 2 cases:
The first case is A > 0, we'll try to arrange the integral above to the form:
$$\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}} = \int \frac{dx}{(mx + n) ^ 2 + \alpha ^ 2}$$
Then use the substitution:
$$mx + n = \alpha \tan t$$

The second case is A < 0, we'll re-arrange it to:
$$\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}} = \int \frac{dx}{\alpha ^ 2 - (mx + n) ^ 2}$$
Then use the substitution:
$$mx + n = \alpha sin t$$

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Is it clear? Can you complete the problem now? :)
Your problem is in the second case.

5. Apr 30, 2007

### hhkeep21

Well I'm having difficulty with the parts problem now because you must make the denominator equal to zero to solve for A and B to solve the equation, but e^x + 1 never equals zero so I'm a little confused how to solve it???

6. May 1, 2007

### Gib Z

$$\frac{1}{u^2-1}=\frac{1}{2(u-1)}-\frac{1}{2(u+1)}$$

Does that help?