# Integrate 1/(9-x^2)dx - Basic Integral Help

• jspek9
In summary, the problem is to integrate 1/(9-x^2)dx. The conversation discusses different methods such as trig substitution and partial fractions. The expert suggests using a trig substitution of u=asinθ and solving for x in terms of θ. However, the initial attempt was incorrect and the expert explains the correct substitution and simplification. The conversation ends with the expert providing helpful insight and suggesting to look up the correct form for future reference.
jspek9
the problem is integrate 1/(9-x^2)dx.

For some reason I'm just drawing a blank. I've tried to fit arccot, ln, and trig sub and get nowhere. Is it part frac decomp? I don't know i think I'm just making it harder than it is.
any insight is appreciated.

Let's see what you've tried for a trig sub (that's how I'd do it).

Partial fractions would also work.

Yes it would, but I didn't want to torture him by demanding that he type out 2 attempts. So I picked my favorite method. :D

I'm using u= asinθ so sinθ= x/3 & dx = 3cosθ dθ & 3cosθ = 9-x2

so subing in I get the integral 3cosθ/3cosθ dθ which reduces to one and equates to x after integration.

The thing is this is really an improper integral problem
Integral from 0 to 3 of 1/(9-x2)
so if I get x when I integrate I have

lim [x]a0
a-->3-

so the lim is 3 but this doesn't seem right

jspek9 said:
I'm using u= asinθ so sinθ= x/3 & dx = 3cosθ dθ

Sure, that's just what I would have done.

& 3cosθ = 9-x2

No, that's not right. It should be the following:

$$3\cos(\theta)=\sqrt{9-x^2}$$

This implies the following.

$$9\cos^2(\theta)=9-x^2$$

Do you see why?

oooooooohhhhh it doesn't fit the form the way i was trying to use it. i probably should have looked up the form before i got so frustrated. oh well ill know next time thanks for the help:)

## 1. What is the formula for integrating 1/(9-x^2)?

The formula for integrating 1/(9-x^2) is:
∫1/(9-x^2)dx = (1/6)ln|3+x| - (1/6)ln|3-x| + C

## 2. Can I use substitution to solve this integral?

Yes, you can use substitution to solve this integral. Let u = 3-x, then du = -dx. Substituting this into the integral, we get:

∫1/(9-x^2)dx = -∫1/u^2du = -(1/u) + C = -(1/(3-x)) + C = (1/6)ln|3-x| + C

## 3. What is the domain of the function 1/(9-x^2)?

The domain of the function 1/(9-x^2) is all real numbers except for x = 3 and x = -3, since these values would make the denominator 0.

## 4. Can I use partial fractions to solve this integral?

Yes, you can use partial fractions to solve this integral. You will need to use the decomposition: 1/(9-x^2) = A/(3+x) + B/(3-x), where A and B are constants. This will allow you to split the integral into two separate integrals that can be easily solved.

## 5. How do I know if I have solved the integral correctly?

You can check your solution by taking the derivative of the integrated function. If the derivative equals the original function, then you have solved the integral correctly. In this case, the derivative of (1/6)ln|3+x| - (1/6)ln|3-x| + C is indeed equal to 1/(9-x^2).

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