Indefinite Integration

• Oct 27th 2010, 08:00 PM
Glitch
Indefinite Integration
The question:

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

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 PM
pickslides
No substitution required, use the fact that

$\int \frac{a}{a^2 + x^2}~dx = Tan^{-1}\left(\frac{x}{a}\right)+C$
• Oct 27th 2010, 09:45 PM
Glitch
Quote:

Originally Posted by pickslides
No substitution required, use the fact that

$\int \frac{a}{a^2 + x^2}~dx = Tan^{-1}\left(\frac{x}{a}\right)+C$

What is this relationship called? I'm a little confused.
• Oct 27th 2010, 09:51 PM
TheEmptySet
This is just a formula for the indefinite integral

here is where it comes from

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

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

$\displaystyle \int \frac{a}{a^2\tan^2(\theta)+a^2}a\sec^2{\theta}d\th eta=\int d\theta =\theta=\tan^{-1}\left(\frac{x}{a} \right)+c$
• Oct 27th 2010, 09:58 PM
Glitch
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 $a^2$
• Oct 28th 2010, 01:52 AM
TheCoffeeMachine
Or $\displaystyle \int \frac{1}{x^2+a^2}\;{dx} = \frac{1}{a}\tan^{-1}{\frac{x}{a}}+k$ (same thing).

Or letting $x = 2\tan{x}$ in the OP's integral (same thing again).
• Oct 28th 2010, 01:08 PM
pickslides
Quote:

Originally Posted by Glitch
Ahh, ok. Is there other similar formulas I should be aware of?

There are quite a few.

Here's some that are similar

$\int \frac{1}{\sqrt{a^2-x^2}}~dx = Sin^{-1}\left(\frac{x}{a}\right)+C$

$\int \frac{-1}{\sqrt{a^2-x^2}}~dx = Cos^{-1}\left(\frac{x}{a}\right)+C$
• Oct 28th 2010, 04:21 PM
TheCoffeeMachine
Quote:

Originally Posted by Glitch
Also, I don't understand how the numerator is considered 'a' in my question when the denominator has '4', which is not $a^2$

It doesn't matter since we can write any number $k$ as $(\sqrt{k})^2$. For example, $4 = (\sqrt{4})^2 = 2^2$.