How do you integrate #1/((1+x^2)^2)#?

Question

Paisley Johnson · Accepted Answer

#x/(2+2x^2)+arctanx/2+C# #I=intdx/(1+x^2)^2# We will use the substitution #x=tantheta#, implying that #dx=sec^2thetad theta#: #I=int(sec^2thetad theta)/(1+tan^2theta)^2# Note that #1+tan^2theta=sec^2theta#: #I=int(sec^2thetad theta)/sec^4theta=int(d theta)/sec^2theta=intcos^2thetad theta# Recall that #cos2theta=2cos^2theta-1#, so #cos^2theta=1/2cos2theta+1/2#. #I=1/2intcos2thetad theta+int1/2d theta# The first integral can be found with substitution (try #u=2theta#). #I=1/4sin2theta+1/2theta+C# From #x=tantheta# we see that #theta=arctanx#. Furthermore, we see that #1/4sin2theta=1/4(2sinthetacostheta)=1/2sinthetacostheta#. Also, since #tantheta=x#, we can draw a right triangle with the side opposite #theta# being #x#, the adjacent side being #1#, and the hypotenuse being #sqrt(1+x^2)#. Thus, #sintheta=x/sqrt(1+x^2)# and #costheta=1/sqrt(1+x^2)#: #I=1/2sinthetacostheta+1/2arctanx+C# #I=1/2(x/sqrt(1+x^2))(1/sqrt(1+x^2))+arctanx/2+C# #I=x/(2(1+x^2))+arctanx/2+C#

Aurora Alvarado · Answer

undefined To integrate 1/((1+x^2)^2), you can use trigonometric substitution. Let $ x = \tan(\theta) $, then $ dx = \sec^2(\theta) d\theta $. Substituting $ x $ and $ dx $ gives:

$$ \int \frac{1}{(1+x^2)^2} dx = \int \frac{1}{(1+\tan^2(\theta))^2} \sec^2(\theta) d\theta $$

$$ = \int \frac{1}{\sec^2(\theta)^2} \sec^2(\theta) d\theta $$

$$ = \int \cos^2(\theta) d\theta $$

Now, use the double-angle identity $ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} $ to integrate:

$$ = \int \frac{1 + \cos(2\theta)}{2} d\theta $$

$$ = \frac{1}{2} \theta + \frac{1}{4} \sin(2\theta) + C $$

Finally, substitute back $ \theta = \arctan(x) $:

$$ = \frac{1}{2} \arctan(x) + \frac{1}{4} \sin(2\arctan(x)) + C $$

This is the integral of $ \frac{1}{(1+x^2)^2} $.