The question:

$\displaystyle \int \frac{dx}{4 + x^2}$

I'm fairly sure I have to use some sort of substitution for this question. But I can't seem to get it to work. >_<

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- Oct 27th 2010, 08:00 PMGlitchIndefinite Integration
**The question:**

$\displaystyle \int \frac{dx}{4 + x^2}$

I'm fairly sure I have to use some sort of substitution for this question. But I can't seem to get it to work. >_< - Oct 27th 2010, 08:07 PMpickslides
No substitution required, use the fact that

$\displaystyle \int \frac{a}{a^2 + x^2}~dx = Tan^{-1}\left(\frac{x}{a}\right)+C$ - Oct 27th 2010, 09:45 PMGlitch
- Oct 27th 2010, 09:51 PMTheEmptySet
This is just a formula for the indefinite integral

here is where it comes from

$\displaystyle \displaystyle \int \frac{a}{x^2+a^2}dx$

let $\displaystyle x=a\tan(\theta) \implies dx=a\sec^2{\theta}d\theta$ this gives

$\displaystyle \displaystyle \int \frac{a}{a^2\tan^2(\theta)+a^2}a\sec^2{\theta}d\th eta=\int d\theta =\theta=\tan^{-1}\left(\frac{x}{a} \right)+c$ - Oct 27th 2010, 09:58 PMGlitch
Ahh, ok. Is there other similar formulas I should be aware of?

Also, I don't understand how the numerator is considered 'a' in my question when the denominator has '4', which is not $\displaystyle a^2$ - Oct 28th 2010, 01:52 AMTheCoffeeMachine
Or $\displaystyle \displaystyle \int \frac{1}{x^2+a^2}\;{dx} = \frac{1}{a}\tan^{-1}{\frac{x}{a}}+k$ (same thing).

Or letting $\displaystyle x = 2\tan{x}$ in the OP's integral (same thing again). - Oct 28th 2010, 01:08 PMpickslides
- Oct 28th 2010, 04:21 PMTheCoffeeMachine