How Do You Integrate 1/(x^2 + 4)?

[nuclearrape66]

how do i integrate 1/(x^2 +4)?

please help

 

[Feldoh]

What's the derivative of arctan?

 

[nuclearrape66]

1/1+x^2

 

[Feldoh]

See how that might be helpful?

 

[rock.freak667]

how do i integrate 1/(x^2 +4)?

please help

$$\int \frac{1}{x^2 +4} dx$$

$$= \int \frac{1}{x^2 +(2)^2} dx$$


try x=2tan$\theta$

 

[nuclearrape66]

hmm lemem see...1 second

 

[rohanprabhu]

1/1+x^2

Exactly.. so now from your function, you take 4 common, you get:

$$\frac{1}{4}\int\frac{1}{({\frac{x}{2}})^2 + 1}dx$$

Now, if you take $\frac{x}{2} = y$.. you can solve this integral.. get a hint?

Once you have done this, it would be helpful for you to remember the formula for a general case as in $$\int\frac{1}{a^2 + x^2}dx$$

 

[Feldoh]

Yeah rohan's post is what I was think too..

 

[nuclearrape66]

oh i see, thanks

and quick response from everyone =)

 

[fermio]

$$\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$
$$\int\frac{1}{x^2+2^2}dx=\frac{1}{2}\arctan\frac{x}{2}+C$$

 

[Feldoh]

$$\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$
$$\int\frac{1}{x^2+2^2}dx=\frac{1}{2}\arctan\frac{x}{2}+C$$

No need to post the solution he all ready figured it out...