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

Question

Emily Ammerman · Accepted Answer

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

I apologize once more. I need to do a self-study review of my calculus before answering any more questions of that kind. Solution verified in Maple.

Samantha Adkison · Answer

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

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 utilize integration by parts to solve the problem as follows (I'll write it out for you if you can't tell; I don't want to write it too long): #=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#

Paisley Johnson · Answer

undefined To find the integral of $\int \frac{dx}{(x^2 + 4)^2}$, use the substitution method with a trigonometric identity:

1. Let $x = 2\tan(\theta)$, implying $dx = 2\sec^2(\theta) d\theta$.
2. Substitute $x = 2\tan(\theta)$ into the integral:
   $$
   \int \frac{2\sec^2(\theta) d\theta}{(4\tan^2(\theta) + 4)^2}
   $$
3. Simplify the denominator using the identity $\tan^2(\theta) + 1 = \sec^2(\theta)$:
   $$
   \int \frac{2\sec^2(\theta) d\theta}{(4\sec^2(\theta))^2} = \int \frac{2\sec^2(\theta) d\theta}{16\sec^4(\theta)}
   $$
4. Simplify further:
   $$
   \int \frac{1}{8} \sec^{-2}(\theta) d\theta = \frac{1}{8} \int \cos^2(\theta) d\theta
   $$
5. Use the half-angle identity: $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$:
   $$
   \frac{1}{8} \int \frac{1 + \cos(2\theta)}{2} d\theta
   $$
6. Separate the integral and integrate:
   $$
   \frac{1}{16} \int (1 + \cos(2\theta)) d\theta = \frac{1}{16} \left( \theta + \frac{\sin(2\theta)}{2} \right) + C
   $$
7. Substitute back for $\theta$: Recall $x = 2\tan(\theta)$, so $\theta = \tan^{-1}\left(\frac{x}{2}\right)$. Also, $\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$ can be substituted as $\frac{2\tan(\theta)}{\sec^2(\theta)}$ or $\frac{2\frac{x}{2}}{1 + \left(\frac{x}{2}\right)^2}$, giving $\frac{x}{x^2/4 + 1}$ after simplification.
8. So, the integral becomes:
   $$
   \frac{1}{16} \left( \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{x^2 + 4} \right) + C
   $$
Therefore, the integral of $\int \frac{dx}{(x^2 + 4)^2}$ is $\frac{1}{16} \left( \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{x^2 + 4} \right) + C$.