Question

How do you find the integral of `int dx / (x²+4)²`?

Answer 1

I gave the wrong answer. Sorry!!! It has been too long since I did any Calculus. The correct answer is:`x/(8(x^2+4)) + 1/16 arctan(x/2)`

My original post deleted

Explanation:

Solution checked in Maple. I need to do a self study refresher on my calculus before I answer any more of that type of question. Sorry again!!!

Answer 2

I found: `1/16{arctan(x/2)+1/2sin{2arctan(x/2)]}+c` but PLEASE check my maths!!!!

Explanation:

I would try to manipulate the denominator setting
`x=2t` so that `dx=2dt`:

`int(2dt)/(4t^2+4)^2=int(2dt)/(4^2(t^2+1)^2)=1/8intdt/(t^2+1)^2=`

now let us set `t=tan(u)` so that `dt=1/cos^2(u)du`
and:

`=1/8int1/(tan^2(u)+1)^2*1/cos^2(u)du=`
`=1/8int1/[((sin^2(u)+cos^2(u))/cos^2(u))]^2*1/cos^2(u)du=`
`=1/8int(cos^4(u))/1*1/cos^2(u)du==1/8intcos^2(u)du=`

here we can use integration by parts and solve it as (if you cannot tell me, I'll write it for you, I do not want to make it too long here):

`=1/8*1/2(u+1/2sin(2u))+c=`

but `u=arctan(t)` so:

`=1/16{arctan(t)+1/2sin[2arctan(t)]}+c=`

and `t=x/2` so finally...

`=1/16{arctan(x/2)+1/2sin{2arctan(x/2)]}+c`