Integrate f(x) = 1/(1+x^2)^2?

Question

I know the answer, just not how to get there. I must be missing something stupidly easy.

Answer 1

Northstar, you dropped the square on the denominator between your second and third steps.

The substitution tan u = x is what you want to use here. This gives you dx = sec² u du. Substituting:

∫1/(1+tan² u)² sec² u du
∫1/(sec² u)² sec² u du
∫sec² u/sec⁴ u du
∫1/sec² u du
∫cos² u du

Using cos² u = (cos (2u) + 1)/2

1/2 + ∫cos (2u) + 1 du
Now we integrate:
sin (2u)/4 + u + C

The back-substitution of u is obvious. Handling the sin (2u)/4 requires:

sin (2u)/4
sin u cos u/2
tan u cos² u/2
tan u/(2 sec² u)
tan u/(2 tan² u + 2)
x/(2x²+2)

So the final answer is:

x/(2x²+2) + arctan (x) + C

Answer 2

Well I wouldn't say stupidly easy but have you been studying substitution integrals recently?

Do you know any trigonometric identities of the form:
1+x^2 = SOMETHING.

Maybe 1 + tan^2(theta) or equivalent. And don't forget the dx term. If you substitute in for x or x^2 you must include the dx substitution.

Answer 3

Let x = tan theta. Then dx = sec^2 theta dtheta

int 1/(1+x^2)^2 dx = int 1/sec^4 theta sec^2 theta dtheta
= int 1/sec^2 theta dtheta
= int cos^2 theta dtheta
= int (1 + cos 2 theta)/2) dtheta
= theta/2 + sin (2 theta) /4 + c
= (arctan x) /2 + x / (2(x^2 + 1)) + c

Answer 4

let x=tan y
dx= (sec^2 y) dy

(sec^2 y)/(1 + tan^2 y)^2=(sec^2 y)/(sec^4 y)=1/sec^2y=cos^2 y =(1 + cos 2y)/2 dy

ans= 1/2y + 1/4sin2y + C

Answer 5

To Terry Scott it extremely is surely cheating thank you to declare/earn some one's admiration or understand or ,of direction,to help. you are able to save visting yahoo solutions and be taught from the solutions published via experts of subjects.i'm specific in case you put in your attempt you would be waiting to place up solutions as appropriate as those given via "outstanding individuals" in this dialogue board. i could desire to quote my adventure and adventure in this dialogue board.After months of careful fact , I truthfully have learnt the methodolgy to comprehend issues.I truthfully have gained education and help each and every time I truthfully have did no longer comprehend.staying power and incessant attempt helped me comprehend fairly some extreme math.in this type, i owe lots to outstanding individuals like hemant sir & many others.So do no longer hasten and have a look into to undertake such skill which might finally finally end up self-defeating. and that i will assure which you would be waiting to delight in and luxuriate in that fulfillment and admiration greater desirable than that earned in one in each and every of those disrespectful way. So wait and notice and be taught.it extremely is thrilled and worthwhile.

Answer 6

Use a trig substitution.

Let
tan θ = x
sec²θ dθ = dx

Now we can integrate.

∫{1/(1+x²)²}dx = ∫{sec²θ/(1 + tan²θ)}dθ = ∫{sec²θ/sec²θ}dθ
= ∫dθ = θ + C = arctan(x) + C